Question

Consider the function f(x)=−2x3+2x2−2x+2 Find the average slope of this function on the interval (4,9). By the Mean Value Theorem, we know there exists a c in the open interval (4,9) such that f′(c) is equal to this mean slope. Find the value of c in the interval which works

Answer 1

Average slope =-242

c=6.67

Explanation:

We are given that a function

`f(x)=-2x^3+2 x^2-2 x+2` and interval (4,9)

We have to find the average slope of the given function and value of c in the given interval.

Using mean value theorem

`{f(b)-f(a)}{b-a}=f'(c)`

a=4 and b=9

`f(9)=-2(9)^3+2(9)^2-2(9)+2`

f(9)=-1312

`f(4)=-2(4)^3+2(4)^2-2(4)+2`

`f(4)=-128+32-8+2=-102`

Substitute the values then we get

`f'(c)=(f(9)-f(4))/(9-4)=(-1312+102)/(5)=-242`

Hence, the average value of slope is -242.

We know that f'(c)=-242

`f'(x)=-6x^2+4x-2`

Substitute x=c

Then
`f'(c)=-6c^2+4c-2`

`-6 c^2+4 c-2=-242`

`-3 c^2+ 2c -1=-121`

Dividing on both sides by 2

`3c^2-2c+1-121=0`

`3c^2-2 c-120=0`

`3c^2-20c+18c-120=0`

`(3c-20)(c+6)=0`

`3c=20` and
`c+6=0`

`c=(20)/(3)` and c=-6 It is not possible because it does not lie in the given interval

Therefore,c=6.67 lie in the given interval