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) =...

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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)` .