`f(x) = e^-x, [0,2]` Find the number `c` that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function, the secant line through the endpoints, and the...

`f(x) = e^-x, [0,2]` Find the number `c` that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function, the secant line through the endpoints, and the tangent line at (c,f(c)). Are the secant line and the tangent line parallel?

Asked on by enotes

Textbook Question

Chapter 4, 4.2 - Problem 14 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

marizi's profile pic

marizi | High School Teacher | (Level 1) Associate Educator

Posted on

Mean Value Theorem states a function f(x) that the satisfies the following hypotheses:

1. a function f(x) that is continuous on the closed interval [a,b] 

2. differentiable on the open interval (a,b)

Then there is a number “c” such that   `a ltcltb` and

 `f'(c) =(f(b)-f(a))/(b-a)`

or   f(b) – f(a) = f’(c) (b-a).

 Mean Value Theorem indicates  that at least one number “c” will exists within the said function. To be able to solve the value of “c”, we consider the special case of Mean Value theorem where we assumed f(a)= f(b)  (Rolle’s Theorem).

 A. For the given function f(x) = `e ^(-x)`  with closed interval [0,2], we solve first for the endpoints.

Plug-in x= 0 in `f(x)=e^(-x)` :

 f(0) = `e^(-0)`

       =`e^(0) `

       =1

 First endpoint: (a, f(a)) = (0,1)

Plug-in x= 2 in f(x)=e^(-x):

 f(0) = `e^(-2) `

         = `(1)/(e^ (2)) `  or 0.135335

  then (b, f(b)) = (2,`(1)/(e^(2))` )

  Second endpoint: (b, f(b)) = (2,`(1)/(e^(2))` )

Law of Exponent:` x^(-n) = (1)/(x^(n))`

B. Solve for the slope of the secant line using the formula:

f'(c) = `(f(b)-f(a))/(b-a) `    

Plug-in the two endpoints:

f'(c) = `((1)/(e^(2))-1)/(2-0) `

f'(c) =  `((1)/(e^(2))-1)/(2)`   

f'(c) =  `(1)/(2e^(2)) - (1)/(2)` or -0.432332358

C. Solve for the derivative function f'(x).

f'(x) =  `e^(-x)` * (-1 dx)       

        = `-e^(-x)` dx

D. Solve for (c, f(c))

Recall the slope of the tangent line = f'(c).

Equate the answers from part B and C:

`((1)/(e^(2))-1)/(2)` =` ` `-e^(-x)`  

  `((1)/(e^(2))-1)/(2)= -(1)/(e^(x))`         

Multiply both sides by -1:

`(((1)/(e^(2))-1)/(2))*(-1) = ((-1)/(e^(x))*(-1)`

`(1-(1)/(e^(2)))/(2)= (1)/(e^(x))`

Cross-multiply to isolate  ` e^(x)` :

`e^(x) = (2)/(1-(1)/(e^(2)))`

Take LN from both sides:

`ln(e^(x)) = ln ((2)/(1-(1)/(e^(2))))`

`x = ln(2) - ln(1-(1)/(e^(2)))`

`x = ln(2) - ln((e^(2)-(1))/(e^(2)))`

`x = ln(2) - (ln(e^(2)-1)-ln(e^(2)))`

`x = ln(2) - ln(e^(2)-1) + ln(e^(2))`

`x = ln(2) - ln(e^(2)-1) + ln(e^(2))`

`x = ln((2)/(e^(2)-1)) + 2`

or x= 0.8385606384 rounded off to c= x =0.8386.

Solve for f(c):

Plug-in x=0.8386 in f(x) =` e^(-x)`

f(0.8386) =` e^(-0.8386)`

               = 0.4323

  Then (c, f(c)) =( 0.8386 , 0.4323)

E. Find tangent line equation.

slope of tangent line = f'(c) =-0.4323

point (c, f(c)) =( 0.8386 , 0.4323)

It shows that secant line and tangent line are parallel.

Using the formula: y=mx+b to solve for b.

0.4323 = (-0.4323) *(0.8386) +b

0.4323 =-0.3625 +b

0.4323 +0.3625 = -0.3625 +0.3625 +b

b=0.7948

Tangent line equation: y = -0.4323x +0.7948

Please see the attached file for the graph.

Blue line is the secant line using the two endpoints: (a, f(a)) and (b, f(b)).

Green line is the tangent line using line equation: y = -0.4323x +0.7948.

Red curve is the graph of `f(x) = e^(-x)` .

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Sources:

We’ve answered 318,972 questions. We can answer yours, too.

Ask a question