Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

Answer 1

Answer:

$x = 1$ satisfies the conclusion of the Mean Value Theorem.

Step-by-step explanation:

According to the Mean Value Theorem, a function must be continuous on $[a,b]$ and differentiable on $(a,b)$. Let $f(x) = 3\cdot x^{2}-2\cdot x +1$ for $[0, 2]$. The function is continuous as domain of polynomial functions is the set of all real numbers.

Now, we proceed to prove that such function is differentiable, that is, that the function has a derivative for all value of $x$ by applying definition of differentiation:

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ (Eq. 1)

Where $f'(x)$ is the derivative of $f(x)$.

Then, we proceed to substitute all values and simplifying the resulting expression by Algebra:

$f'(x) = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-2\cdot (x+h)+1-3\cdot x^{2}+2\cdot x -1}{h}$

$f'(x) = \lim_{h \to 0} \frac{3\cdot x^{2}+6\cdot x\cdot h + 3\cdot h^{2}-2\cdot x -2\cdot h +1 -3\cdot x^{2}+2\cdot x - 1}{h}$

$f'(x) = \lim_{h \to 0} \frac{6\cdot x\cdot h +3\cdot h^{2}-2\cdot h}{h}$

$f'(x) = \lim_{h \to 0} 6\cdot x+ \lim_{h \to 0} 3\cdot h - \lim_{h \to 0} 2$

$f'(x) = 6\cdot x -2$

Which means that the derivative of the quadratic function is a linear function, whose domain is the set of all real numbers. Hence, $f(x)$ is differentiable and satisfies all hypotheses.

Given that $f(x)$ is continuous and differentiable, the following condition must be satisfied:

$f'(c) = \frac{f(b)-f(a)}{b-a}$, $c\in[a,b]$ (Eq. 2)

If $a = 0$ and $b = 2$, then function evaluted at each bound is, respectively:

$f(0) = 3\cdot (0)^{2}-2\cdot (0) +1$

$f(0) = 1$

$f(2) = 3\cdot (2)^{2}-2\cdot (2) +1$

$f(2) = 9$

By replacing each term on (Eq. 2), we get this expanded version:

$6\cdot c -2 = \frac{9-1}{2-0}$

$6\cdot c -2 = 4$

$c = 1$

$x = 1$ satisfies the conclusion of the Mean Value Theorem.