Question

does the function satisfy the hypotheses of the mean value theorem on the given interval?
$f(x)=4x^{2}-3x + 2$, $0,2$
o yes, it does not matter if $f$ is continuous or differentiable, every function satisfies the mean value theorem.
o yes, $f$ is continuous on $0,2$ and differentiable on $(0,2)$ since polynomials are continuous and differentiable on $mathbb{r}$.
o no, $f$ is not continuous on $0,2$.
o no, $f$ is continuous on $0,2$ but not differentiable on $(0,2)$.
o there is not enough information to verify if this function satisfies the mean value theorem.
if it satisfies the hypotheses, find all numbers $c$ that satisfy the conclusion of the mean value theorem. (enter your answers as a comma - separated list. if it does not satisfy the hypotheses, enter dne)
$c=$

Explanation:

Step1: Check continuity and differentiability

Since \(f(x)=4x^{2}-3x + 2\) is a polynomial function. Polynomial functions are continuous on \(\mathbb{R}\), so \(f(x)\) is continuous on \([0,2]\). Polynomial functions are differentiable on \(\mathbb{R}\), so \(f(x)\) is differentiable on \((0,2)\).

Step2: Apply the Mean Value Theorem formula

The Mean Value Theorem states that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\), where \(a = 0\), \(b=2\).
First, find \(f(0)\) and \(f(2)\):
\(f(0)=4\times0^{2}-3\times0 + 2=2\)
\(f(2)=4\times2^{2}-3\times2 + 2=16-6 + 2=12\)
Then \(\frac{f(2)-f(0)}{2-0}=\frac{12 - 2}{2}=5\)
Next, find the derivative \(f^{\prime}(x)\):
Using the power rule \((x^{n})^{\prime}=nx^{n - 1}\), \(f^{\prime}(x)=8x-3\)
Set \(f^{\prime}(c)=8c-3\) equal to \(5\) (from \(\frac{f(2)-f(0)}{2-0}\)):
\(8c-3 = 5\)

Step3: Solve for \(c\)

Add \(3\) to both sides of the equation \(8c-3 = 5\):
\(8c=5 + 3=8\)
Divide both sides by \(8\):
\(c = 1\)

Answer:

Yes, \(f\) is continuous on \([0,2]\) and differentiable on \((0,2)\) since polynomials are continuous and differentiable on \(\mathbb{R}\). \(c = 1\)