Respuesta :
Answer:
only one number c=0 in the interval [-1,1]
Step-by-step explanation:
Given : Function [tex]f(x) = |(x^2-9)(x^2 + 1)|[/tex] 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 [tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
Solution : f(x) is a function that satisfies all of the following :
1) f(x) is continuous on the closed interval [-1,1]
[tex]\lim_{x\to a} f(x)=f(a)[/tex]
2) f(x) is differentiable on the open interval (-1,1)
Then there is a number c such that [tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
[tex]f(a)=f(-1) = |(-1^2-9)(-1^2 + 1)|=|(-8)(2)|=16[/tex]
[tex]f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16[/tex]
[tex]f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0[/tex]
[tex]f'(c)=0[/tex] ........[1]
Now, we find f'(x)
[tex]f(x) = |(x^2-9)(x^2 + 1)|[/tex]
[tex]f(x) =x^4-8x^2-9[/tex]
Differentiating w.r.t x
[tex]f'(x) =4x^3-16x[/tex]
In place of x we put x=c
[tex]f'(c) =4c^3-16c[/tex]
[tex]f'(c) =4c^3-16c=0[/tex] (by [1], f'(c)=0)
[tex]4c(c^2-4)=0[/tex]
[tex]4c=0,c^2-4=0[/tex]
either c=0 or [tex]c^2-4=0\rightarrow c=\pm2[/tex]
we cannot take [tex]c=\pm2[/tex] 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.
