3.1 Definition of derivatives; Questions 1-10

[Eigenvalue]

For the following equations, find the slope of the secant line between values [tex]x_{1 }[/tex] and [tex]x_{2}[/tex]

1. f(x)=4x+7; [tex]x_{1 }[/tex]=2, [tex]x_{2}[/tex]=5
2. f(x)=8x-3; [tex]x_{1 }[/tex]=-1, [tex]x_{2}[/tex]=3
3. f(x)=x²+2x+1; [tex]x_{1 }[/tex]=3, [tex]x_{2}[/tex]=3.5
4. f(x)=-x²+x+2; [tex]x_{1 }[/tex]=0.5, [tex]x_{2}[/tex]=1.5
5. f(x)=[tex]\frac{4}{3x-1}[/tex]; [tex]x_{1 }[/tex]=1, [tex]x_{2}[/tex]=3

6. f(x)=[tex]\frac{x-7}{2x+1}[/tex]; [tex]x_{1 }[/tex]=0, [tex]x_{2}[/tex]=2

7. f(x)=[tex]\sqrt{x}[/tex]; [tex]x_{1 }[/tex]=1, [tex]x_{2}[/tex]=16
8. f(x)=[tex]\sqrt{x-9}[/tex]; [tex]x_{1 }[/tex]=10, [tex]x_{2}[/tex]=13

9. f(x)=[tex]x^{1/3}[/tex]; [tex]x_{1 }[/tex]=0, [tex]x_{2}[/tex]=8

10.f(x)=6[tex]x^{2/3 }[/tex]+2[tex]x^{1/3 }[/tex]; [tex]x_{1 }[/tex]=1, [tex]x_{2}[/tex]=27

[Eigenvalue]

1. f(x)=4x+7; [tex]x_{1 }[/tex]=2, [tex]x_{2}[/tex]=5

a) Substitute in 2 and 5 for x to find the y-values for when x is 2 and x is 5:
f(2)=4(2)+7=15
f(5)=4(5)+7=27

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{27-15}{5-2}[/tex]=[tex]\frac{12}{3}[/tex]=4

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-27=4(x-5)
y=4x+7

d) Check using the other point:
15=4(2)+7
15=15 ✓

e) Conclusion: the secant line for [tex]x_{1 }[/tex] and [tex]x_{2}[/tex]--where [tex]x_{1 }[/tex] and [tex]x_{2 }[/tex] are x-values of points on a linear function--is always the line itself, as a line connects two points

[Eigenvalue]

2. f(x)=8x-3; [tex]x_{1 }[/tex]=-1, [tex]x_{2}[/tex]=3

a) Substitute in -1 and 3 for x to find the y-values for when x is -1 and x is 3:
f(2)=8(-1)-3=-11
f(5)=8(3)-3=21

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{21-(-11)}{3-(-1)}[/tex]=[tex]\frac{32}{4}[/tex]=8

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y+11=8(x+1)
y=8x-3

d) Check using the other point:
21=8(3)-3
21=21 ✓

e) Conclusion: the secant line for [tex]x_{1 }[/tex] and [tex]x_{2}[/tex]--where [tex]x_{1 }[/tex] and [tex]x_{2 }[/tex] are x-values of points on a linear function--is always the line itself, as a line connects two points

[Eigenvalue]

3. f(x)=x²+2x+1; [tex]x_{1 }[/tex]=3, [tex]x_{2 }[/tex]=3.5

a) Substitute in 3 and 3.5 for x to find the y-values for when x is 3 and x is 3.5:
f(3)=(3+1)²=16
f(3.5)=(3.5+1)²=(4.5)²=20.25

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{20.25-16}{3-3.5}[/tex]=[tex]\frac{4.25}{0.5}[/tex]=8.5

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-20.25=8.5(x-3.5)
y=8.5x-9.5

d) Check using the other point:
16=8.5(3)-9.5
16=25.5-9.5
16=16 ✓

[Eigenvalue]

4. f(x)=-x²+x+2 [tex]x_{1 }[/tex]=0.5, [tex]x_{2 }[/tex]=1.5

a) Substitute in 0.5 and 1.5 for x to find the y-values for when x is 0 and x is 1.5:
f(0.5)=-(1.5)²+1.5+2=3.5-2.25=1.25
f(1.5)=-(0.5)²+0.5+2=2.5-0.25=2.25

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{2.25-1.25}{0.5-1.5}[/tex]=[tex]\frac{-1}{1}[/tex]=-1

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-1.5=-(x-1.25)
y=-x+2.75

d) Check using the other point:
2.25=-(0.5)+2.75
2.25=2.25 ✓

[Eigenvalue]

7. f(x)=[tex]\sqrt{x}[/tex]; [tex]x_{1 }[/tex]=1, [tex]x_{2 }[/tex]=16

a) Substitute in 1 and 16 for x to find the y-values for when x is 1 and x is 16:
f(1)=[tex]\sqrt{1}[/tex]=1
f(16)=[tex]\sqrt{16}[/tex]=4

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{4-1}{16-1}[/tex]=[tex]\frac{3}{15}[/tex]=0.2

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-4=0.2(x-16)
y=0.2x+0.8

d) Check using the other point:
1=0.2(1)+0.8
1=1 ✓

[Eigenvalue]

8. f(x)=[tex]\sqrt{x-9}[/tex]; [tex]x_{1 }[/tex]=10, [tex]x_{2 }[/tex]=13

a) Substitute in 10 and 13 for x to find the y-values for when x is 10 and x is 13:
f(10)=[tex]\sqrt{10-9}[/tex]=1
f(13)=[tex]\sqrt{13-9}[/tex]=2

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{2-1}{13-10}[/tex]=[tex]\frac{1}{3}[/tex]

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-2=[tex]\frac{1}{3}[/tex](x-13)
y=[tex]\frac{1}{3}[/tex]x-[tex]\frac{7}{3}[/tex]

d) Check using the other point:
1=[tex]\frac{1}{3}[/tex](10)-[tex]\frac{7}{3}[/tex]
1=1 ✓

[Eigenvalue]

9. f(x)=[tex]x^{1/3 }[/tex]; [tex]x_{1 }[/tex]=0, [tex]x_{2 }[/tex]=8

a) Substitute in 0 and 8 for x to find the y-values for when x is 0 and x is 8:
f(0)=[tex]\sqrt[3]{0}[/tex]=0
f(8)=[tex]\sqrt[3]{8}[/tex]=2

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{2-0}{8-0}[/tex]=[tex]\frac{2}{8}[/tex]=0.25

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-2=0.25(x-8)
y=0.25x

d) Check using the other point:
0=0(0.25)
0=0 ✓

[Eigenvalue]

10. f(x)=6[tex]x^{2/3}[/tex]+2[tex]x^{1/3 }[/tex]; [tex]x_{1 }[/tex]=1, [tex]x_{2 }[/tex]=27

a) Substitute in 1 and 27 for x to find the y-values for when x is 1 and x is 27:
f(1)=6([tex]1^{2/3 }[/tex])+2([tex]1^{1/3 }[/tex])=6+2=8
f(27)=6[tex]x^{2/3}[/tex]+2([tex]x^{1/3}[/tex])=6(9)+2(3)=60

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{60-8}{27-1}[/tex]=[tex]\frac{52}{26}[/tex]=2

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-60=2(x-27)
y=2x+6

d) Check using the other point:
8=2(1)+6
8=8 ✓

[Eigenvalue]

5. f(x)=[tex]\frac{4}{3x-1}[/tex]; [tex]x_{1 }[/tex]=1, [tex]x_{2 }[/tex]=3

a) Substitute in 1 and 3 for x to find the y-values for when x is 1 and x is 3:
f(1)=[tex]\frac{4}{3(1)-1}[/tex]=[tex]\frac{4}{2}[/tex]=2
f(3)=[tex]\frac{4}{3(3)-1}[/tex]=[tex]\frac{4}{8}[/tex]=0.5

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{0.5-2}{3-1}[/tex]=[tex]\frac{-1.5}{2}[/tex]=-0.75

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y-0.5=-0.75(x-3)
y=-0.75x+2.65

d) Check using the other point:
2=-0.75(1)+2.75
2=2.75-0.75
2=2 ✓

[Eigenvalue]

6. f(x)=[tex]\frac{x-7}{2x+1}[/tex]; [tex]x_{1 }[/tex]=0, [tex]x_{2 }[/tex]=2

a) Substitute in 0 and 2 for x to find the y-values for when x is 0 and x is 2:
f(0)=[tex]\frac{0-7}{2(0)+1}[/tex]=-7
f(2)=[tex]\frac{2-7}{2(2)+1}[/tex]=[tex]\frac{-5}{5}[/tex]=-1

b) Find the slope by finding the quotient of the change in y-values and the change in x-values:
m=[tex]\frac{∆y}{∆x}[/tex]=[tex]\frac{-1-(-7)}{2-0}[/tex]=[tex]\frac{6}{2}[/tex]=3

c) Find the equation of the line using point-slope form:
y-[tex]y_{1 }[/tex]=m(x-[tex]x_{1 }[/tex])
y+7=3x
y=3x-7

d) Check using the other point:
-1=3(2)-7
-1=6-7
-1=-1 ✓