Question

A function f that satisfies the hypothesis of the Mean

Value Theorem on [0, 4] and for which there are exactly

three values c ∈ (0, 4) that satisfy the conclusion of the

theorem .

Short answer

The function$f\left(x\right)=\sqrt{x}$satisfied the conclusion of Mean value theorem .

Step-by-step solution

Step 1. Given information .

Consider the function $f\left(x\right)=\sqrt{x}$satisfied the conditions of Mean value theorem on$\left[0,4\right]$.

Step 2. Using Mean value theorem .

If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one value c ∈ (a, b) such that ,

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

Step 3. Classifying the theorem for function fx=x .

The given function$f\left(x\right)=\sqrt{x}$is continuous on closed interval $\left[0,4\right]$and differentiable on $\left(0,4\right)$therefore it satisfied the condition of Mean value theorem .

$$\begin{gathered} f\left(x\right)=\sqrt{x} \\ f\text{'}\left(x\right)=\frac{1}{2}{x}^{1/2} \end{gathered}$$

Further simplify .

$f\text{'}\left(c\right)=1$

At point $\left[0,4\right]$the value of c is 1 therefore the function satisfied all conditions .

Step 4. Plot the graph .

The graph of the given function is shown below .