Question

Find the number c guaranteed by the mean value theorem for derivatives for f(x)=(x+1)^3, [-1,1]

Answer 1

Solution. Using mean value theorem: Suppose f(x) is a function that satisfies all of the following.

1) f(x) is continuous on the closed interval [a;b].

2) f(x) is differentiable on the open interval (a;b).

Then there is a number c such that a<c<b and

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

Function  f(x)=(x+1)^3 is continuous on the closed interval [-1,1] and f(x) is differentiable on the open interval (-1,1).

$$f(-1)=(-1+1)^3=0$$

$$f(1)=(1+1)^3=8$$

Therefore

$$f'(c)=\frac {8-0} {1-(-1)}=4$$

Find the first derivative of the function f(x).

$$f'(x)=3(x+1)^2$$

As result get equation

$$3(x+1)^2=4$$

Find the roots of the equation.

$$x+1=\frac {2} {\sqrt {3}} \Longrightarrow x_1=-1+\frac {2} {\sqrt {3}}$$

$$x_1 \isin (-1,1)$$

$$x+1=-\frac {2} {\sqrt {3}} \Longrightarrow x_2=-1-\frac {2} {\sqrt {3}}$$

$$x_2 \notin (-1,1)$$

Answer.

$$-1+\frac {2} {\sqrt {3}}$$