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

Answer:

Average slope =-242

c=6.67

Step-by-step 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

$f'(c)=\frac{f(9)-f(4)}{9-4}=\frac{-1312+102}{5}=-242$

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 $c+6=0$

$c=\frac{20}{3}$ 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