Applications of Derivatives: Problems with Solutions

Solution:
$y=\frac{x^{3}}{4}-3x\Longrightarrow y'=\frac{3}{4}x^{2}-3$

We solve $y'=0$:

$\frac{3}{4}x^{2}=3\Longrightarrow x^{2}=4\Longrightarrow x=\pm 2$

The intervals of monotonicity are $\left( -\infty ,-2\right),\ \left( -2,2\right),\ \left( 2,\infty \right)$

We evaluate the sign of the derivative at a point inside each interval:

$y'(-3)=\frac{3}{4}(-3)^{2}-3=\frac{27}{4}-3>0$, so the function is increasing on $\left( -\infty ,-2\right)$

$y'(0)=\frac{3}{4}(0)^{2}-3=-3<0$, so the function is decreasing on $\left( -2,2\right)$

$y'(3)=\frac{3}{4}(3)^{2}-3=\frac{27}{4}-3>0$, so the function is increasing on $\left( 2,\infty \right)$

Answer: increasing on $\left( -\infty ,-2\right)$ and $\left( 2,\infty \right)$, decreasing on $\left( -2,2\right)$

Solution:
$f(x)=x^{4}-2x^{2}\Longrightarrow f'(x)=4x^{3}-4x=0$
We 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 evaluate $f'(a)$, where $a$ is a value taken from each interval:

$f'(-2)=4(-2)^{3}-4(-2)=-32+8<0$, so $f$ is decreasing on $(-\infty ,-1)$

$f'(-\frac{1}{2})=4\left( -\frac{1}{2}\right)^{3}-4\left( -\frac{1}{2}\right) =-\frac{1}{2}+2>0$, so $f$ is increasing on $(-1,0)$

$f'(\frac{1}{2})=4\left( \frac{1}{2}\right)^{3}-4\left( \frac{1}{2}\right) =\frac{1}{2}-2<0$, so $f$ is decreasing on $(0,1)$

$f'(2)=4(2)^{3}-4(2)=32-8>0$, so $f$ is increasing on $(1,\infty )$

Statement (i) is false, because $f$ changes from increasing to decreasing at $x=0$.

Answer: (ii), (iii), (iv) are true.

Solution:
$h(x)=27x-x^{3}\Longrightarrow h'(x)=27-3x^{2}=0\Longrightarrow x=\pm 3$

The intervals of monotonicity are $\left( -\infty ,-3\right),\ \left( -3,3\right),\ \left( 3,\infty \right)$

$h'(-4)=27-3\left( -4\right)^{2}=27-48<0\Longrightarrow$ 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 $\left( 3,\infty \right)$

Answer: increasing on $\left( -3,3\right)$ and decreasing on $\left( -\infty ,-3\right) \cup \left( 3,\infty \right)$

Solution:
$y=x+\frac{4}{x}\Longrightarrow y'=1-\frac{4}{x^{2}}=\frac{x^{2}-4}{x^{2}}$

$y'=0$ when $x^{2}-4=0$, that is $x=\pm 2$. The point $x=0$ is not in the domain of the function, but it also splits the number line.

The intervals of monotonicity are $\left( -\infty ,-2\right),\ \left( -2,0\right),\ \left( 0,2\right),\ \left( 2,\infty \right)$

$y'(-3)=\frac{9-4}{9}>0$, so the function is increasing on $\left( -\infty ,-2\right)$

$y'(-1)=\frac{1-4}{1}<0$, so the function is decreasing on $\left( -2,0\right)$

$y'(1)=\frac{1-4}{1}<0$, so the function is decreasing on $\left( 0,2\right)$

$y'(3)=\frac{9-4}{9}>0$, so the function is increasing on $\left( 2,\infty \right)$

Answer: increasing on $\left( -\infty ,-2\right) \cup \left( 2,\infty \right)$ and decreasing on $\left( -2,0\right) \cup \left( 0,2\right)$

Solution:
First we find the critical points:

$f'(x)=2\left( x-1\right) \left( x+3\right) +\left( x-1\right)^{2}$

$f'(x)=2x^{2}+4x-6+x^{2}-2x+1=3x^{2}+2x-5=0$

$x=\frac{-2\pm \sqrt{4+60}}{6}=\frac{-2\pm 8}{6}\Longrightarrow x=1$ and $x=-\frac{5}{3}$

$f(1)=0$ and $f\left( -\frac{5}{3}\right) =\left( -\frac{8}{3}\right)^{2}\cdot \frac{4}{3}=\frac{256}{27}$, so the critical points are $\left( 1,0\right)$ and $\left( -\frac{5}{3},\frac{256}{27}\right)$, and statement a) is true.

Now we check the sign of $f'$ on each interval:

$f'(-2)=12-4-5=3>0$, so $f$ is increasing on $\left( -\infty ,-\frac{5}{3}\right)$

$f'(0)=-5<0$, so $f$ is decreasing on $\left( -\frac{5}{3},1\right)$

$f'(2)=12+4-5=11>0$, so $f$ is increasing on $\left( 1,\infty \right)$

This proves b). The derivative changes from $+$ to $-$ at $x=-\frac{5}{3}$ and from $-$ to $+$ at $x=1$, so there is a maximum at $\left( -\frac{5}{3},\frac{256}{27}\right)$ and a minimum at $\left( 1,0\right)$, which proves c).

Answer: all are true.

Solution:
$f(x)=x^{4}-32x+4\Longrightarrow f'(x)=4x^{3}-32=0$

$4x^{3}=32\Longrightarrow x^{3}=8\Longrightarrow x=2$

So the only critical point is $x=2$, not $x=4$: statement a) is false.

$f'(0)=-32<0$, so the function is decreasing on $\left( -\infty ,2\right)$ and statement b) is false.

$f'(3)=108-32>0$, so the function is increasing on $\left( 2,\infty \right)$. The derivative changes from $-$ to $+$ at $x=2$, therefore there is a minimum at $x=2$ and statement c) is true.

Answer: only c) is true.

Solution:
$f(x)=\left( x+2\right)^{2/3}\Longrightarrow f'(x)=\frac{2}{3}\left( x+2\right)^{-1/3}=\frac{2}{3\sqrt[3]{x+2}}$

The derivative is never equal to zero, but it does not exist at $x=-2$. Since $f(-2)=0$, the only critical point is $\left( -2,0\right)$ and statement a) is false.

For $x<-2$ we have $\sqrt[3]{x+2}<0$, so $f'(x)<0$ and the function is decreasing on $\left( -\infty ,-2\right)$

For $x>-2$ we have $\sqrt[3]{x+2}>0$, so $f'(x)>0$ and the function is increasing on $\left( -2,\infty \right)$

This is exactly the opposite of statement b), so b) is false.

The derivative changes from $-$ to $+$ at $x=-2$, so the point $\left( -2,0\right)$ is a minimum, not a maximum, and c) is false.

Answer: all are false.

Solution:
We differentiate twice:

$f(x)=\frac{1}{4}x^{4}-2x^{2}\Longrightarrow f'(x)=x^{3}-4x\Longrightarrow f''(x)=3x^{2}-4$

$f''(x)=0\Longrightarrow x=\pm \frac{2\sqrt{3}}{3}$

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 $f''(c)$ at a point $c$ from each interval:

$f''(-2)=3\cdot 4-4=8>0\Longrightarrow f$ 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 $\left( \frac{2\sqrt{3}}{3},\infty \right)$

$f\left( \pm \frac{2\sqrt{3}}{3}\right) =\frac{1}{4}\cdot \frac{16}{9}-2\cdot \frac{4}{3}=\frac{4}{9}-\frac{8}{3}=-\frac{20}{9}$

At both points the second derivative changes sign, so the inflection points are $\left( -\frac{2\sqrt{3}}{3},-\frac{20}{9}\right)$ and $\left( \frac{2\sqrt{3}}{3},-\frac{20}{9}\right)$

Answer: C

Solution:
We differentiate twice:

$f(x)=2x^{4}-8x+3\Longrightarrow f'(x)=8x^{3}-8\Longrightarrow f''(x)=24x^{2}$

The second derivative vanishes only at $x=0$, but it does not change sign there, so there is no inflection point.

$f''(x)=24x^{2}\geq 0$ so the function is convex on $\left( -\infty ,\infty \right)$.

Answer: A

Solution:
We differentiate twice:

$y=-x^{3}+3x^{2}-2\Longrightarrow y'=-3x^{2}+6x\Longrightarrow y''=-6x+6$

$y''=0\Longrightarrow x=1$

The intervals of concavity are $\left( -\infty ,1\right)$, $\left( 1,\infty \right)$

We check the sign of the second derivative:

$y''(0)=6>0$, so the function is convex on $\left( -\infty ,1\right)$

$y''(2)=-6<0$, so the function is concave on $\left( 1,\infty \right)$

The second derivative changes sign at $x=1$.

Answer: B

Solution:
We differentiate twice:

$f(x)=-x^{3}+6x^{2}-9x-1\Longrightarrow f'(x)=-3x^{2}+12x-9\Longrightarrow f''(x)=-6x+12$

$f''(x)=0\Longrightarrow x=2$

The intervals of concavity are $\left( -\infty ,2\right)$, $\left( 2,\infty \right)$

We check the sign of the second derivative:

$f''(0)=12>0$, so the function is convex on $\left( -\infty ,2\right)$

$f''(3)=-6<0$, so the function is concave on $\left( 2,\infty \right)$

The second derivative changes sign at $x=2$.

Answer: D

Solution:
We find the critical points:

$f'(x)=4x^{3}-12x^{2}=4x^{2}(x-3)=0$ so $x=0$ and $x=3$

The second derivative is $f''(x)=12x^{2}-24x=12x(x-2)$

At $x=3$ we have $f''(3)=36>0$, so there is a minimum at $\left( 3,-25\right)$.

At $x=0$ the derivative $f'(x)=4x^{2}(x-3)$ does not change sign, so there is no extremum there.

$f(3)=81-108+2=-25$

$f''$ changes sign at $x=0$ and at $x=2$, therefore the inflection points are $\left( 0,2\right)$ and $\left( 2,-14\right)$.

$f''(-1)=36>0,\ f''(1)=-12<0,\ f''(3)=36>0$, so $f$ is convex on $\left( -\infty ,0\right) \cup \left( 2,\infty \right)$ and concave on $\left( 0,2\right)$.

Answer: A

Solution:
The derivative is $f'(x)=\frac{2}{3}x^{-1/3}=\frac{2}{3\sqrt[3]{x}}$

It is never equal to zero and it does not exist at $x=0$, so the only critical point is $\left( 0,-3\right)$.

For $x<0$ we have $f'(x)<0$ and for $x>0$ we have $f'(x)>0$, so the function decreases and then increases: there is a minimum at $\left( 0,-3\right)$.

The second derivative is $f''(x)=-\frac{2}{9}x^{-4/3}=-\frac{2}{9\sqrt[3]{x^{4}}}$

It is negative for every $x\neq 0$, so the graph is concave on $\left( -\infty ,0\right)$ and on $\left( 0,\infty \right)$, and there is no inflection point.

Answer: C

Solution:
Using the chain rule:

$f'(x)=2\left( \left( x^{2}+3\right)^{5}+x\right) \cdot \left( 5\left( x^{2}+3\right)^{4}\cdot 2x+1\right)$

$f'(x)=2\left( \left( x^{2}+3\right)^{5}+x\right) \left( 10x\left( x^{2}+3\right)^{4}+1\right)$

Now we substitute $x=-1$:

$\left( -1\right)^{2}+3=4,\ 4^{5}=1024,\ 4^{4}=256$

$f'(-1)=2\left( 1024-1\right) \left( 10\cdot \left( -1\right) \cdot 256+1\right) =2\cdot 1023\cdot \left( -2559\right)$

$f'(-1)=-5235714$

Answer: $f'(-1)=-5235714$

Solution:
Using the chain rule:

$f'(x)=\frac{1}{2\sqrt{2+\sqrt{2+\sqrt{x}}}}\cdot \frac{1}{2\sqrt{2+\sqrt{x}}}\cdot \frac{1}{2\sqrt{x}}$

Now we substitute $x=4$:

$\sqrt{4}=2,\ \sqrt{2+2}=2,\ \sqrt{2+2}=2$

$f'(4)=\frac{1}{2\cdot 2}\cdot \frac{1}{2\cdot 2}\cdot \frac{1}{2\cdot 2}=\frac{1}{4}\cdot \frac{1}{4}\cdot \frac{1}{4}=\frac{1}{64}$

Answer: $f'(4)=\frac{1}{64}$

Solution:
We differentiate:

$f(x)=\sqrt{25-x^{2}}\Longrightarrow f'(x)=\frac{-2x}{2\sqrt{25-x^{2}}}=\frac{-x}{\sqrt{25-x^{2}}}$

The slope of the tangent line at $x=3$ is $f'(3)=\frac{-3}{\sqrt{25-9}}=\frac{-3}{4}$

The equation of the tangent line through the point $(3,4)$ is:

$y-4=-\frac{3}{4}\left( x-3\right)$

$4y-16=-3x+9$

$4y+3x=25$

Answer: A

Solution:
When $t=\frac{\pi }{8}$ we have $12t=\frac{3\pi }{2}$, so $\cos \frac{3\pi }{2}=0$, $\sin \frac{3\pi }{2}=-1$

The position is $y=\frac{1}{3}\cdot 0-\frac{1}{4}\cdot \left( -1\right) =\frac{1}{4}$ feet

We differentiate to obtain the velocity:

$y'=-\frac{12}{3}\sin 12t-\frac{12}{4}\cos 12t=-4\sin 12t-3\cos 12t$

$y'\left( \frac{\pi }{8}\right) =-4\cdot \left( -1\right) -3\cdot 0=4$, so the velocity is $4$ ft/sec

Answer: A

Solution:
Since $\left\vert \cos 8t\right\vert \leq 1$, the largest value of $\theta $ is $0.2$ radians, which is reached when $\cos 8t=1$.

We differentiate with respect to the time: $\theta =0.2\cos 8t\Longrightarrow \frac{d\theta }{dt}=-1.6\sin 8t$

The rate of change when $t=3$ seconds is:

$\frac{d\theta }{dt}(3)=-1.6\sin \left( 8\cdot 3\right) =-1.6\sin 24\approx -1.6\cdot \left( -0.9056\right) \approx 1.449$ rad/sec

(the angle $8\cdot 3=24$ radians is measured in radians, not in degrees)

Answer: C