Question

Determine whether the Mean Value Theorem can be applied to f on the closed interval [a, b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = f (b) − f (a) b − a . If the Mean Value Theorem cannot be applied, explain why not.
f(x) = x^3 + 2x, [-1,1]

Answer 1

Answer:

There is a value of $c$ in (-1, 1), $c = 0.577$.

Step-by-step explanation:

Let $f(x) = x^{3}+2\cdot x$ for $x \in[-1,1]$, we need to prove that $f(x)$ is continuous and differentiable to apply the Mean Value Theorem. Given that $f(x)$ is a polynomical function, its domain comprises all real numbers and therefore, function is continuous.

If $f(x)$ is differentiable, then $f'(x)$ exists for all value of $x$. By definition of derivative, we obtain the following expression:

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{(x+h)^{3}+2\cdot (x+h)-x^{3}-2\cdot x}{h}$

$f'(x) = \lim_{h \to 0} \frac{x^{3}+3\cdot x^{2}\cdot h+3\cdot x\cdot h^{2}+h^{3}+2\cdot x+2\cdot h-x^{3}-2\cdot x}{h}$

$f'(x) = \lim_{h \to 0} \frac{3\cdot x^{2}\cdot h+3\cdot x\cdot h^{2}+h^{3}+2\cdot h}{h}$

$f'(x) = \lim_{h \to 0} 3\cdot x^{2}+ \lim_{h \to 0} 3\cdot x \cdot h+ \lim_{h \to 0} h^{2}+ \lim_{h \to 0} 2$

$f'(x) = 3\cdot x^{2}+2$ (Eq. 2)

The derivative of a cubic function is quadratic function, which is also a polynomic function. Hence, the function is differentiable at the given interval.

According to the Mean Value Theorem, the following relationship is fulfilled:

$f'(c) = \frac{f(1)-f(-1)}{1-(-1)}$ (Eq. 3)

If we know that $f(-1) = -3$, $f(1) = 3$ and $f'(c) = 3\cdot c^{2}+2$, then we expand the definition as follows:

$3\cdot c^{2}+2 = 3$

$3\cdot c^{2} = 1$

$c = \sqrt{\frac{1}{3} }$

$c \approx 0.577$

There is a value of $c$ in the interval (-1, 1), $c = 0.577$.