If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that  `f'(c)=(f(b)-f(a))/(b-a)`  .   1.   `f(x) =sqrt(1-x)`        [-8,1]

1 Answer | Add Yours

Top Answer

lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

First, we have to determine if the function is continuous on the given interval [-8,1]. To do so, let's determine the domain of f(x). Since it is a radical function, we have to refer to the properties of square root. In square root, negative numbers are not allowed inside it. So, set the radicand greater than or equal to zero.

`1-x>=0` 

`-x>=-1`

`x<=1`

Hence, the domain of f(x) is `(-oo,1]` .

Since the given interval falls within the domain of f(x), then

`f(x)=sqrt(1-x)`

is continuous on the interval [-8,1].

Thus, we can apply the Mean Value Theorem to solve for c on the open interval (-8,1).

To solve for c, apply the formula

`f'(c)=(f(b)-f(a))/(b-a)`

Take note that the right side refers to the slope of the line connecting the end values of the interval. This line is called as secant line.

And for the left side, it is the slope of the line that is tangent to the graph of f(x) at x=c.

So, let's solve for the slope of these two lines.

For the secant line, its slope is:

`m_(SL)=(f(b)-f(a))/(b-a)`

`m_SL=(sqrt(1-1)-sqrt(1-(-8)))/(1-(-8))`

`m_(SL)=(sqrt0-sqrt9)/9`

` ` `m_(SL)=-3/9`

`m_(SL)=-1/3`

For the tangent line, take the derivative of the function.

`m_(TL)=f'(x)`

`m_(TL)=1/2(1-x)^(-1/2)*(-1)`

`m_(TL)=-1/(2sqrt(1-x))`

Since it is tangent at x=c, plug-in this to the derivative.

`m_(TL)=-1/(2sqrt(1-c))`

Then, set the slope of each line equal to each other.

`-1/(2sqrt(1-c))=-1/3`

And isolate c.

`1/(2sqrt(1-c))=1/3`

`3=2sqrt(1-c)`

`3^2=(2sqrt(1-c))^2`

`9=4(1-c)`

`9/4=1-c`

`9/4-1=-c`

`5/4=-c`

`-5/4=c`

So, on the open interval (-8,1), the slope of the secant line connecting the end values of the interval is parallel to the tangent at x=-5/4. Thus, `c=-5/4`   .

We’ve answered 318,917 questions. We can answer yours, too.

Ask a question