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Practice
Applications of Derivatives
Applications of Derivatives: Problems with Solutions
By
Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
In the following graph estimate the open intervals over which the function is increasing or decreasing.
$y=\frac{x^{3}}{4}-3x$
Apply the first derivative criterium.
Decreasing on $\left( -\infty ,-2\right)$, $\left( -2,2\right)$ and $\left( 2,\infty \right)$
Increasing on $\left( -\infty ,-2\right)$, decreasing on $\left( -2,2\right)$ and $\left( 2,\infty \right)$
Increasing on $\left( -\infty ,-2\right) $ and $\left( 2,\infty \right)$, decreasing on $\left( -2,2\right)$
Decreasing on $\left( -\infty ,-2\right)$ and $\left( 2,\infty \right)$, increasing on $\left( -2,2\right)$
Solution:
Answer: The function is increasing on $\left( -\infty ,-2\right) $ and $\left( 2,\infty \right)$, decreasing on $\left( -2,2\right) $
$y=\frac{x^{3}}{4}-3x\Longrightarrow y'=\frac{3}{4}x^{2}-3=0$ and we solve this equation, then
$\frac{3}{4}x^{2}=3\Longrightarrow x^{2}=4$
$x=\pm 2$
The intervals of monotonicity are:
$\left( -\infty ,-2\right);\ \left( -2,2\right);\ \left( 2,\infty \right) $
We must evaluate the sign of the derivative at a point within each interval, so
$y'(-3);\ y'(0);\ y'(3)$
$y'(-3)=\frac{3}{4}(-3)^{2}-3=\frac{27}{4}-3>0$
then the function is increasing on $\left( -\infty ,-2\right) $
$y'(0)=\frac{3}{4}(0)^{2}-3=0-3<0$ then the function is decreasing on $\left( -2,2\right) $
$y'(3)=\frac{3}{4}(3)^{2}-3=\frac{27}{4}-3>0$
then the function is increasing on $\left( 2,\infty \right) $
Answer:
increasing on $\left( -\infty ,-2\right) $ and $\left( 2,\infty \right)$, decreasing on $\left( -2,2\right) $
Problem 2
Look at the graph of the function $f(x)=x^{4}-2x^{2}$
What can you say about the following propositions involving its graph?
(i) It is increasing on $(-1,1)$
(ii) It is decreasing on $(-\infty ,-1)$
(iii) It is increasing on $(-1,0)$
(iv) It is decreasing on $(0,1)$
(ii), (iii), (iv) are true.
Only (i) is true.
(ii),(iv) are false.
All are false.
Solution:
Answer: (ii),(iii),(iv) are true
$f(x)=x^{4}-2x^{2}\Longrightarrow f'(x)=4x^{3}-4x=0$
Solve the equation:
$4x^{3}-4x=4x(x^{2}-1)=0\Longrightarrow x=0,-1,1$
The intervals of monotonicity are:
$(-\infty ,-1);(-1,0);(0,1);(1,\infty )$
We must evaluate $f'(a)$ where $a$ is a value from each interval.
$f'(-2)=4\left( -2\right)^{3}-4\left( -2\right)=-32+8<0$ then $f$ is decreasing on $(-\infty ,-1)$
$f'(-\frac{1}{2})=4\left( -\frac{1}{2}\right) ^{3}-4\left( -\frac{1}{2}\right) =-\frac{1}{2}+2>0$ then $f$ is increasing on $(-1,0)$
$f'(\frac{1}{2})=4\left( \frac{1}{2}\right)^{3}-4\left( \frac{1}{2}\right) =\frac{1}{2}-2<0$ then $f$ is decreasing on $(0,1)$
$f'(2)=4\left( 2\right)^{3}-4\left( 2\right) =32-8>0$
then $f$ is increasing on $(1,\infty )$
Problem 3
Identify the open intervals over which the function $h(x)=27x-x^{3}$ is increasing or decreasing.
The function is increasing
The function is decreasing
Increasing on $\left( -3,3\right) $ and decreasing on $\left( -\infty ,-3\right) \cup \left( 3,\infty \right) $
Decreasing on $\left( -3,3\right) $ and increasing on $\left( -\infty ,-3\right) \cup \left( 3,\infty \right) $
Solution:
$h(x)=27x-x^{3}\Longrightarrow $
$h'(x)=27-3x^{2}=0\Longrightarrow x=\pm 3$
The intervals of monotonicity are $\left( -\infty ,-3\right);\ \left( -3,3\right);\ \left( 3,\infty \right)$
Then $h'(-4)=27-3\left( -4\right)^{2}=27-48<0\Longrightarrow $ the function is decreasing on $\left( -\infty ,-3\right) $
$h'(0)=27-3\left( 0\right)^{2}=27>0\Longrightarrow$ the function is increasing on $\left( -3,3\right) $
$h'(4)=27-3\left( 4\right)^{2}=27-48<0\Longrightarrow$ the function is decreasing on $\left( 3,\infty \right)$
Problem 4
Find the derivative of the function $y=x+\frac{4}{x}$. Determine when the function increases and decreases.
Increasing on $\left( -\infty ,-2\right) \cup (2,\infty )$ and decreasing on $\left( -2,0\right) \cup (0,2)$
Increasing on $\left( -\infty ,-2\right) $ and decreasing on $(2,\infty )$
Decreasing on $\left( -\infty ,-2\right) \cup (2,\infty )$ and increasing on $\left( -2,0\right) \cup (0,2)$
Decreasing on $\left( -\infty ,-2\right) $ and increasing on $(2,\infty )$
Solution:
$y=x+\frac{4}{x}\Longrightarrow y'=1-\frac{4}{x^{2}}=\frac{x^{2}-4}{x^{2}}=0$ or $x^{2}=0$
We obtain the following solutions $x=\pm2,0$
The intervals of monotonicity are:
$\left(-\infty ,-2\right);\ \left( -2,0\right);\ \left( 0,2\right);\ \left( 2,\infty\right) $
We must analize each interval.
$\frac{dy}{dx}(-3)=\frac{\left( -3\right)^{2}-4}{\left( -3\right)^{2}}=\frac{5}{9}>0$
$y$ is increasing on $\left(-\infty ,-2\right) $
$\frac{dy}{dx}(-1)=\frac{\left( -1\right)^{2}-4}{\left( -1\right)^{2}}=\frac{-3}{1}<0$
$y$ is decreasing on $\left( -2,0\right) $
$\frac{dy}{dx}(1)=\frac{\left( 1\right)^{2}-4}{\left( 1\right)^{2}}=\frac{-3}{1}<0$
$y$ is decreasing on $\left(0,2\right) $
$\frac{dy}{dx}(3)=\frac{\left( 3\right)^{2}-4}{\left( 3\right)^{2}}=\frac{5}{9}>0$
$y$ is increasing on $\left(2,\infty \right) $
Problem 5
Given the function $f(x)=\left( x-1\right)^{2}\left( x+3\right)$
a) The critical points of $f$ are $\left( 1,0\right) ;\left( -\frac{5}{3},\frac{256}{27}\right) $
b) The function is increasing on $\left( -\infty ,-\frac{5}{3}\right) \cup \left( 1,\infty \right) $ and decreasing on $\left( -\frac{5}{3},1\right) $
c) The function has a maximun at $\left( -\frac{5}{3},\frac{256}{27}\right)$ and has a minimum at $\left(1,0\right)$
A and b are true.
Only c is true.
All are true.
All are false.
Solution:
Answer: All are true.
First we need to find the critical points $\frac{d}{dx}f(x)=2\left( x-1\right) \left( x+3\right) +\left( x-1\right)^{2}$
$\frac{d}{dx}f(x)=2x^{2}+4x-6+x^{2}-2x+1=3x^{2}+2x-5=0$
$x=\frac{-2\pm \sqrt{4+60}}{6}=\frac{-2\pm 8}{6}=1$ and $-\frac{5}{3}$
The critical points of $f$ are $x=1$ $f(1)=0\Longrightarrow \left( 1,0\right)$
and $x=-\frac{5}{3}$ $f(-\frac{5}{3})=\left( -\frac{5}{3}-1\right)^{2}\left( -\frac{5}{3}+3\right) =\frac{256}{27}\Longrightarrow \left( -\frac{5}{3},\frac{256}{27}\right) $
Analize the intervals $\left( -\infty ,-\frac{5}{3}\right);\ \left( -\frac{5}{3},1\right);\ \left( 1,\infty \right)$
$\frac{d}{dx}f(-2)=3(-2)^{2}+2(-2)-5= 3\Longrightarrow $increasing on $\left( -\infty ,-\frac{5}{3} \right) $
$\frac{d}{dx}f(0)=3(0)^{2}+2(0)-5= -5\Longrightarrow $decreasing on $\left( -\frac{5}{3},1\right) $
$\frac{d}{dx}f(2)=3(2)^{2}+2(2)-5=11\Longrightarrow $increasing on $\left( 1,\infty \right) $
Now we need to find the second derivative of the function $f$
$\frac{d^{2}}{dx^{2}}f(x)=\frac{d}{dx}(3x^{2}+2x-5)=6x+2$
Now evaluate at each critical point: $\frac{d^{2}}{dx^{2}}f(1)=8>0\Longrightarrow $
There exists a minimum at $\left( 1,0\right) $
$\frac{d^{2}}{dx^{2}}f(-\frac{5}{3})=6(-\frac{5}{3})+2= -8<0\Longrightarrow $
There exist a maximum at $\left( -\frac{5}{3},\frac{256}{27}\right)$
Problem 6
Let $f(x)$ be defined as $f(x)=x^{4}-32x+4$
a) The critical point of $f$ is at $x=4$ only.
b) The function is increasing on $\left(-\infty, 2\right)$
c) The function has a minimum at $x=2$
A and b are true.
Only c is true.
All are true.
All are false.
Solution:
Answer: Only c is true.
$\frac{d}{dx}f(x)=4x^{3}-32=0\Longrightarrow x=2;f(2)=2^{4}-32(2)+4=-44$
The function has one critical point - $\left( 2,-44\right)$ and
intervals of monotonicity - $\left( -\infty ,2\right);\ (2,\infty )$
$\frac{d}{dx}f(1)=4(1)^{3}-32<0 \Longrightarrow $ decreasing on $\left( -\infty,2\right)$
$\frac{d}{dx}f(3)=4(3)^{3}-32>0 \Longrightarrow $increasing on $\left( 2,\infty \right) $
$\frac{d^{2}}{dx^{2}}f(x)=\frac{d}{dx}(4x^{3}-32)=12x^{2}$
$\frac{d^{2}}{dx^{2}}f(2)=12(2)^{2}=48>0 \Longrightarrow $
There are a minimum at $\left( 2,-44\right)$
Problem 7
Graph the function $f(x)=\left( x+2\right)^{2/3}$
a) The critical point of $f$ is $(0,0)$ only.
b) The function is increasing on $\left( -\infty ,-2\right)$ and and decreasing on $(-2,\infty )$
c) The function has a maximun at $\left(-2,0\right) $
A and b are true.
Only c is true.
All are true.
All are false.
Solution:
Answer: All are false.
$f(x)=\left( x+2\right) ^{2/3}\Longrightarrow \frac{d}{dx}f(x)=\frac{2}{3}\left( x+2\right)^{-1/3}=\frac{2}{3\sqrt[3]{x+2}}$
We obtain the critical point when we do $\sqrt[3]{x+2}=0$
then
$x=-2\Longrightarrow \left( -2,f(-2)\right) =\left( -2,0\right) $ is the critical point
Now we define the intervals of monotonicity $\left( -\infty,-2\right);\ (-2,\infty)$
Since $\frac{d}{dx}f(-3)=\frac{2}{3\sqrt[3]{-3+2}}=-\frac{2}{3}<0$ then $f$ is decresing on $\left( -\infty ,-2\right)$ and $\frac{d}{dx}f(0)=\frac{2}{3\sqrt[3]{0+2}}>0$ then $f$ is incresing on $(-2,\infty )\Longrightarrow $
There is a minimum at point $\left( -2,0\right) $
Problem 8
Find the inflection points of the function $f(x)=\frac{1}{4}x^{4}-2x^{2}$ and analyze the concavity of the function
A) Concave: $\left( \frac{2\sqrt{3}}{3},\infty \right)$, inflection point: $\left( \frac{2\sqrt{3}}{3},-\frac{20}{9}\right)$
B) Convex: $\left( \frac{2\sqrt{3}}{3},\infty \right)$, inflection point: $\left( \frac{2\sqrt{3}}{3},-\frac{20}{9}\right) $
C) Convex: $\left( -\infty ,-\frac{2\sqrt{3}}{3}\right) \cup \left( \frac{2\sqrt{3}}{3},\infty \right)$, concave on: $\left( -\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}\right)$, inflection points: $\left( -\frac{2\sqrt{3}}{3},-\frac{20}{9}\right) ;\left( \frac{2\sqrt{3}}{3},-\frac{20}{9}\right)$
D) Concave: $\left( -\infty ,-\frac{2\sqrt{3}}{3}\right) \cup \left( \frac{2\sqrt{3}}{3},\infty \right)$, convex on: $\left( -\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}\right)$, inflection points: $\left( -\frac{2\sqrt{3}}{3},-\frac{20}{9}\right) ;\left( \frac{2\sqrt{3}}{3},-\frac{20}{9}\right)$
A
B
C
D
Solution:
Answer: C
$f(x)=\frac{1}{4}x^{4}-2x^{2}\Longrightarrow f'(x)=x^{3}-4x$
$\frac{d^{2}}{dx^{2}}f(x)=3x^{2}-4=0\Longrightarrow x=\pm \frac{2\sqrt{3}}{3}$
The intervals of concavity are $\left( -\infty ,-\frac{2\sqrt{3}}{3}\right);\ \left( -\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}\right);\ \left( \frac{2\sqrt{3}}{3},\infty \right) $ Now we evaluate $f''(c)$ where $c$ is a point from each interval.
$f''(-2)=8>0\Longrightarrow f$ is convex on $\left( -\infty ,-\frac{2\sqrt{3}}{3}\right)$
$f''(0)=-4<0\Longrightarrow f$ is concave on $\left( -\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3}\right)$
$f''(2)=8>0\Longrightarrow f$ is convex on $\left( \frac{2\sqrt{3}}{3},\infty \right) $
Then, there are two points of inflection $\left( -\frac{2\sqrt{3}}{3},f(-\frac{2\sqrt{3}}{3})\right);\left( \frac{2 \sqrt{3}}{3},f(\frac{2\sqrt{3}}{3})\right)$
$f(-\frac{2\sqrt{3}}{3})=f(\frac{2\sqrt{3}}{3})=\frac{1}{4}(\frac{2\sqrt{3}}{3})^{4}-2(\frac{2\sqrt{3}}{3})^{2}=(\frac{2\sqrt{3}}{3})^{2}\left[ \frac{1}{4}(\frac{2\sqrt{3}}{3})^{2}-2\right] $
$f(-\frac{2\sqrt{3}}{3})=f(\frac{2\sqrt{3}}{3})=\frac{4}{3}\left( \frac{1}{4}\cdot \frac{4}{3}-2\right) =\allowbreak -\frac{20}{9}$
So, the points of inflection are $\left( -\frac{2\sqrt{3}}{3},-\frac{20}{9}\right);\left( \frac{2\sqrt{3}}{3},-\frac{20}{9}\right) $
Problem 9
Consider the following function $f(x)=2x^{4}-8x+3$
Find the points of inflection and analyze the concavity.
A) Convex on $\left( -\infty ,\infty\right) $ and there are no points of inflection.
B) Convex on $\left( -\infty ,0\right)$, concave on $\left( 0,\infty \right)$, a points of inflection at $\left( 0,3\right)$
C) Concave on $\left( -\infty ,0\right)$, convex on $\left( 0,\infty \right)$, a point of inflection at $\left( 0,3\right)$ D) Concave on $\left( -\infty ,\infty \right)$, there are no points of inflection.
A
B
C
D
Solution:
$f(x)=2x^{4}-8x+3\Longrightarrow \frac{d}{dx}f(x)=8x^{3}-4$
$\frac{d^{2}}{dx^{2}}f(x)=24x^{2}=0\Longrightarrow x=0$, then the intervals of concavity are $\left( -\infty ,0\right);\ \left( 0,\infty \right)$
Now we evaluate $\frac{d^{2}}{dx^{2}}f(c)$ where $c$ is a point on each interval. $\frac{d^{2}}{dx^{2}}f(-1)=24(-1)^{2}>0 \Longrightarrow f$ is convex on $\left( -\infty ,0\right) $
$\frac{d^{2}}{dx^{2}}f(1)=24(1)^{2}>0 \Longrightarrow f$ is convex on $\left( 0,\infty \right)$
Since there are no changes in concavity, we conclude that the function has no points of inflection.
Problem 10
Determine the concavity of $y=-x^{3}+3x^{2}-2$
A) $f$ is concave on $\left( -\infty,1\right)$, convex on $\left( 1,\infty \right) $
B) $f$ is convex on $\left( -\infty,1\right)$, concave on $\left( 1,\infty \right) $
C) $f$ is concave on $\left( -\infty,0\right)$, convex on $\left( 0,\infty \right) $
D) $f$ is convex on $\left( -\infty,2\right)$, concave on $\left( 2,\infty \right) $
A
B
C
D
Solution:
Answer: B.
$y=-x^{3}+3x^{2}-2$ $\Longrightarrow \frac{d}{dx}y=-3x^{2}+6x$
$\frac{d^{2}}{dx^{2}}y=-6x+6=0\Longrightarrow x=1$
The intervals of concavity are $\left( -\infty,1\right);\ \left( 1,\infty \right)$
Now we evaluate $\frac{d^{2}}{dx^{2}}y(c)$
where $c$ is a point on each interval
$\frac{d^{2}}{dx^{2}}y(0)=-6(0)+6>0 \Longrightarrow f$ is convex on $\left( -\infty ,1\right)$
$\frac{d^{2}}{dx^{2}}y(2)=-6(2)+6<0 \Longrightarrow f$ is concave on $\left( 1,\infty \right)$
There are a point of inflection at $(1,0)$
Problem 11
Determine the concavity of $f(x)=-x^{3}+6x^{2}-9x-1$
A) $f$ is concave on $\left( -\infty,1\right)$, convex on $\left( 1,\infty \right)$
B) $f$ is concave on $\left( -\infty,2\right)$, convex on $\left( 2,\infty \right)$
C) $f$ is convex on $\left( -\infty,0\right)$, concave on $\left( 0,\infty \right)$
D) $f$ is convex on $\left( -\infty,2\right)$, concave on $\left( 2,\infty \right)$
A
B
C
D
Solution:
Answer: D
$f(x)=-x^{3}+6x^{2}-9x-1\Longrightarrow \frac{d}{dx}f(x)=-3x^{2}+12x-9$
$\frac{d^{2}}{dx^{2}}f(x)=-6x+12=0\Longrightarrow x=2$
The intervals of concavity are $\left( -\infty,2\right);\ \left( 2,\infty \right)$
Evaluate $f''(c)$ where $c$ is a point on each interval
$f''(0)=-6(0)+12>0 \Longrightarrow f$ is convex on $\left( -\infty ,2\right)$
$f''(3)=-6(3)+12<0 \Longrightarrow f$ is concave on $\left( 2,\infty \right)$
The function has a point of inflection at $(2,-3)$.
Problem 12
Let we have the function $f(x)=x^{4}-4x^{3}+2$
1. Find all relative extremes and inflection points.
2. Use the criterion of the second derivative where appropriate.
A) Minimum at $(3,-25), f$ is convex on $\left( -\infty ,0\right) \cup \left( 2,\infty \right)$, concave on $\left(0,2\right)$
B) Maximum at $(3,-25), f$ is convex on $\left( -\infty ,0\right)$, concave on $\left( 2,\infty \right)$
C) Maximum at $(0,2), f$ is convex on $\left( 0,2\right)$, concave on $\left( 2,\infty \right)$
D) minimum at $(0,2), f$ is convex on $\left( 0,2\right)$, concave on $\left( 2,\infty \right)$
A
B
C
D
Solution:
Answer: A
$f(x)=x^{4}-4x^{3}+2\Longrightarrow \frac{d}{dx}f(x)=4x^{3}-12x^{2}$
$ \frac{d^{2}}{dx^{2}}f(x)=12x^{2}-24x$ then we find the critical points
$\frac{d}{dx}f(x)=4x^{3}-12x^{2}=0\Longrightarrow x^{2}(4x-12)=0\Longrightarrow x=0$ or $x=3$ are the critical points.
Since $\frac{d}{dx}f(-1)=4(-1)^{3}-12(-1)^{2}=-16<0$ and $\frac{d}{dx}f(1)=4(1)^{3}-12(1)^{2}=-8<0$, there are no extreme value in $(0,f(0))$
Now we take $x=4\Longrightarrow \frac{d}{dx}f(1)=4(4)^{3}-12(4)^{2}=256-192>0$
So, we get a minimum point at $(3,f(3))=(3,-25)$
Now we analize the concavity, find the intervals
$\frac{d^{2}}{dx^{2}}f(x)=12x^{2}-24x=0\Longrightarrow x(12x-24)=0\Longrightarrow x=0$ or $x=2$
The concavity intervals are $\left( -\infty ,0\right) ;\left( 0,2\right) ;\left( 2,\infty \right) $ and since $\frac{d^{2}}{dx^{2}}f(-1)=12(-1)^{2}-24(-1)>0\Longrightarrow f$ is convex on $\left( -\infty ,0\right) $
$\frac{d^{2}}{dx^{2}}f(1)= 12(1)^{2}-24(1)<0 \Longrightarrow f$ is concave on $\left( 0,2\right) $
$\frac{d^{2}}{dx^{2}}f(3)=12(3)^{2}-24(3)>0\Longrightarrow f$ is convex on $\left( 2,\infty \right)$
Problem 13
Find all relative extreme and inflection points of the function: $f(x)=x^{2/3}-3$.
A) Maximum at $\left( 0,-3\right);$ concave on $\left( -\infty, 0\right) $
B) Maximum at $\left( 0,-3\right);$ concave on $\left( 0,\infty\right) $
C) Minimum at $\left( 0,-3\right);$ concave on $\left( -\infty,0\right) \cup \left( 0,\infty \right) $
D) Maximum at $\left( 0,-3\right);$ convex on $\left( -\infty,0\right) \cup \left( 0,\infty \right) $
A
B
C
D
Solution:
$f(x)=x^{2/3}-3\Longrightarrow \frac{d}{dx}(x^{2/3}-3)=\frac{2}{3}x^{-1/3}=\frac{2}{3\sqrt[3]{x}}=0$
There is no solution for $x$, but we must solve when the derivative does not exist, so $3\sqrt[3]{x}=0\Longrightarrow x=0$ then there is a critical point at $\left( 0,f(0)\right) =\left( 0,-3\right) $
The intervals of monotonicity are $\left( -\infty ,0\right) ;\left( 0,\infty \right) $
Since $\frac{d}{dx}f(-1)=-\frac{2}{3}$ and $\frac{d}{dx}f(1)=\frac{2}{3}$ then $\left( 0,-3\right) $ is a minimum poin.t
If we find the second derivative $\frac{d}{dx}\left( \frac{2}{3}x^{-1/3}\right) =-\frac{2}{9}x^{-4/3}$ $\frac{d^{2}}{dx^{2}}f(x)=-\frac{2}{9\sqrt[3]{x^{4}}}$ then the concavity intervals are $\left( -\infty ,0\right);\ \left( 0,\infty \right)$ since $\frac{d^{2}}{dx^{2}}f(-1)=-\frac{2}{9}$ the function is concave ($\cap $) on $\left( -\infty ,0\right)$ and since $\frac{d^{2}}{dx^{2}}f(1)=-\frac{2}{9}$ the function is concave ($\cap $) on $\left( 0,\infty \right) $
Problem 14
Let $f(x)=\left( \left( x^{2}+3\right)^{5}+x\right)^{2}$.
Find $\frac{d}{dx}\ f(-1)$
$\frac{d}{dx}\ f(-1)=-1$
$\frac{d}{dx}\ f(-1)=-5235714$
$\frac{d}{dx}\ f(-1)=0$
$\frac{d}{dx}\ f(-1)=123654$
Solution:
$\frac{d}{dx} f(x)=\frac{d}{dx} \left[ \left( \left(x^{2}+3\right)^{5}+x\right)^{2}\right] =2\left( \left( x^{2}+3\right)^{5}+x\right) \cdot \frac{d}{dx}\left( \left( x^{2}+3\right) ^{5}+x\right) $
$\frac{d}{dx}$ $f(x)=2\left( \left( x^{2}+3\right)^{5}+x\right) \cdot \left( 5\left( x^{2}+3\right) ^{4}\cdot \frac{d}{dx}% \left( x^{2}+3\right) +1\right) $
$\frac{d}{dx}$ $f(x)=2\left( \left( x^{2}+3\right)^{5}+x\right) \cdot \left( 5\left( x^{2}+3\right) ^{4}\cdot 2x+1\right) $
$\frac{d}{dx}$ $f(-1)=2\left( \left( (-1)^{2}+3\right)^{5}-1\right) \cdot \left( 5\left( (-1)^{2}+3\right) ^{4}\cdot 2(-1)+1\right) $
$\frac{d}{dx}$ $f(-1)=2(4^{5}-1)\cdot (-10(4)^{4}+1)=-5235714$
Problem 15
Let $f(x)=\sqrt{2+\sqrt{2+\sqrt{x}}}$
Find $\frac{d}{dx}\ f(4)$
$\frac{d}{dx}\ f(4)=\frac{1}{32}$
$\frac{d}{dx}\ f(4)=\frac{1}{16}$
$\frac{d}{dx}\ f(4)=2$
$\frac{d}{dx}\ f(4)=\frac{1}{64}$
Solution:
$\frac{d}{dx}f(x)=\frac{\frac{d}{dx}(2+\sqrt{2+\sqrt{x}})}{2\sqrt{2+\sqrt{2+\sqrt{x}}}}=\frac{\frac{d}{dx}(2+\sqrt{x})}{2\sqrt{2+\sqrt{x}}}\cdot \frac{1}{2\sqrt{2+\sqrt{2+\sqrt{x}}}}=\frac{1}{2\sqrt{x}}\cdot \frac{1}{2\sqrt{2+\sqrt{x}}}\cdot \frac{1}{2\sqrt{2+\sqrt{2+\sqrt{x}}}}$
$\frac{d}{dx}f(x)=\frac{1}{8\sqrt{x}\sqrt{2+\sqrt{x}}\sqrt{2+\sqrt{2+\sqrt{x}}}}$
Now we evaluate at $x=4$
$\frac{d}{dx}f(4)=\frac{1}{8\sqrt{4}\sqrt{2+\sqrt{4}}\sqrt{2+\sqrt{2+\sqrt{4}}}}=\frac{1}{16\cdot 2\cdot 2}=\frac{1}{64}$
Problem 16
Find the equation of the tangent line to the function $f(x)=\sqrt{25-x^{2}}$ at point $(3,4)$
A) $4y+3x=25$ is the tangent line.
B) $4x+3y=25$ is the tangent line.
C) $3y-4x=25$ is the tangent line.
D) None of the above.
A
B
C
D
Solution:
$\frac{d}{dx}f(x)=\frac{-x}{\sqrt{25-x^{2}}}\Longrightarrow m=\frac{d}{dx}f(x)=\frac{-3}{\sqrt{25-3^{2}}}=-\frac{3}{4}$
The equation of tangent line is of the form $y=m(x-x_{0})+y_{0}$
$y=-\frac{3}{4}(x-3)+4=-\frac{3}{4}x+\frac{25}{4}\Longrightarrow y+\frac{3}{4}x=\frac{25}{4}$
$4y+3x=25$ is the tangent line.
Problem 17
The displacement of its equilibrium position for an object in harmonic motion located at the end of a spring is:
$y=\frac{1}{3}\cos 12t-\frac{1}{4}\sin12t$ where $y$ is measured in feet and $t$ in seconds.
Determine the position and speed of the object when $t=\frac{\pi}{8}$.
A) Position: $\frac{1}{4}$feet, speed: $4$ ft/sec
B) Position: $0$ feet, speed: $2$ ft/sec
C) Position: -$\frac{1}{4}$feet, speed: $4$ ft/sec
D) Position: $0$ feet, speed: $-2$ ft/sec
A
B
C
D
Solution:
Answer: A
The speed is the derivative of position function $\frac{d}{dt}y=-4\sin 12t+3\cos 12t$
To find the position we need to do $y(\frac{\pi}{8})=\frac{1}{3}\cos 12\frac{\pi }{8}-\frac{1}{4}\sin12\frac{\pi }{8}$
$y(\frac{\pi}{8})=\frac{1}{3}\cos \frac{3}{2}\pi -\frac{1}{4}\sin\frac{3}{2}\pi =\frac{1}{3}\cdot 0-\frac{1}{4}(-1)=\frac{1}{4}$ feet
The speed when $t=\frac{\pi}{8}$ is $\frac{d}{dt}y(\frac{\pi}{8})=-4\sin 12\frac{\pi }{8}+3\cos 12\frac{\pi }{8}$
$\frac{d}{dt}y(\frac{\pi}{8})=-4\sin \frac{3}{2}\pi +3\cos \frac{3}{2}\pi = -4(-1)+3(0)=4$ ft/sec
Problem 18
A 15 cm pendulum moves according to the equation:
$\theta =0.2\cos 8t,$ where $\theta $ is the angular displacement of the vertical in radians and $t$ is the time in seconds.
Calculate the maximum angular displacement and the reason of change of $\theta $ when $t=3$ seconds.
A) M.A.D $8t$ radians; R.O.C $0.2$rad/sec
B) M.A.D $0.2$ radians; R.O.C $8$ rad/sec
C) M.A.D $0.2$ radians; R.O.C $1.449$ rad/sec
D) none of above
A
B
C
D
Solution:
Answer: C
$\theta =0.2\cos 8t\Longrightarrow \frac{d}{dt}\theta =-1.6\sin 8t$ then the reason of change when $t=3$ seconds
We obtain it, if we do $\frac{d}{dt}\theta (3)=-1.6\sin 8(3)=\allowbreak 1.\,\allowbreak 449$ rad/sec
Now we find the extrem values of $\theta =0.2\cos 8t$
So $\frac{d}{dt}\theta =-1.6\sin 8t=0$
then $8t=n\pi$ with $n=0,1,2,3...\Longrightarrow t=\frac{n\pi }{8}n=0,1,2,3...$ Now we obtain the second derivate of $\theta =0.2\cos 8t\Longrightarrow \frac{d^{2}}{dt^{2}}\theta =-12.8\cos 8t$
So $\frac{d^{2}}{dt^{2}}\theta (\frac{n\pi }{8})=-12.8\cos 8(\frac{n\pi }{8})=-12.8\cos n\pi $ this expression is negative when $n=0,2,4,6...$ since $\cos n\pi >0$
then the maximum angular displacement occur when $t=\frac{n\pi }{8};n=0,2,4..2k$
$\Longrightarrow t=\frac{k\pi }{4};k=0,1,2,3,..$
Now calculate the maximum angular displacement
$\theta (\frac{k\pi }{4})=0.2\cos 8\frac{k\pi }{4}=0.2\cos 2k\pi =0.2$ radians
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