How do I find the numbers $c$ $c$ that satisfy the Mean Value Theorem for $f (x) = x^{3} + x - 1$ $f(x)=x^3+x-1$ on the interval $[0, 3]$ $[0,3]$ ?

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Answer 1 · 2014-09-28T16:29:08

The value of $c$ $c$ is $\sqrt{3}$ $sqrt{3}$ .

Let us look at some details.

M.V.Thm. states that there exists $c$ $c$ in (0,3) such that

$f'(c)={f(3)-f(0)}/{3-0}$ .

Let us find such $c$ .

The left-hand side is

$f'(c)=3c^2+1$ .

The right-hand side is

${f(3)-f(0)}/{3-0}={29-(-1)}/{3}=10$ .

By setting them equal to each other,

$3c^2+1=10 Rightarrow 3x^2=9 Rightarrow x^2=3 Rightarrow x=pm sqrt{3}$

Since $0<c<3$ , $c=sqrt{3}$ .

I hope that this was helpful.

How do I find the numbers ccc that satisfy the Mean Value Theorem for f(x)=x^3+x-1f(x)=x3+x−1f(x)=x^3+x-1 on the interval [0,3][0,3][0,3] ?