Question

If f(x) = |(x2 − 9)(x2 + 1)|, how many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem?

Answer 1

Answer:

only one number c=0 in the interval [-1,1]

Step-by-step explanation:

Given : Function $f(x) = |(x^2-9)(x^2 + 1)|$   in the interval [-1,1]

To find : How many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem.

Mean value theorem : If f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point 'c' in (a,b) such that $f'(c)=\frac{f(b)-f(a)}{b-a}$

Solution : f(x) is a function that satisfies all of the following :

1) f(x)  is continuous on the closed interval [-1,1]  

$\lim_{x\to a} f(x)=f(a)$

2) f(x) is differentiable on the open interval  (-1,1)

Then there is a number  c such that  $f'(c)=\frac{f(b)-f(a)}{b-a}$

$f(a)=f(-1) = |(-1^2-9)(-1^2 + 1)|=|(-8)(2)|=16$

$f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16$

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0$

$f'(c)=0$ ........[1]

Now, we find f'(x)

$f(x) = |(x^2-9)(x^2 + 1)|$

$f(x) =x^4-8x^2-9$

Differentiating w.r.t  x

$f'(x) =4x^3-16x$

In place of x we put x=c

$f'(c) =4c^3-16c$

$f'(c) =4c^3-16c=0$  (by [1], f'(c)=0)

$4c(c^2-4)=0$

$4c=0,c^2-4=0$

either c=0 or  $c^2-4=0\rightarrow c=\pm2$

we cannot take $c=\pm2$  because they don't lie in the interval [-1,1]

Therefore, there is only one number c=0 which lie in interval [-1,1] and satisfying the conclusion of the mean value theorem.