show that (f(x)=\frac{1}{2}x - sqrt{x}) satisfies all assumptions of rolles theorem on (0,4), and find all…

Explanation:

Step1: Check continuity

The function $f(x)=\frac{1}{2}x - \sqrt{x}$ is a sum of a linear - function $\frac{1}{2}x$ and a square - root function $\sqrt{x}$. Both $\frac{1}{2}x$ and $\sqrt{x}$ are continuous on $[0,4]$, so $f(x)$ is continuous on $[0,4]$.

Step2: Check differentiability

Find the derivative of $f(x)$ using the power rule. The derivative of $\frac{1}{2}x$ is $\frac{1}{2}$, and the derivative of $\sqrt{x}=x^{\frac{1}{2}}$ is $\frac{1}{2}x^{-\frac{1}{2}}$. So $f'(x)=\frac{1}{2}-\frac{1}{2\sqrt{x}}$, and $f(x)$ is differentiable on $(0,4)$.

Step3: Check $f(0)$ and $f(4)$

Calculate $f(0)=\frac{1}{2}(0)-\sqrt{0}=0$ and $f(4)=\frac{1}{2}(4)-\sqrt{4}=2 - 2=0$.

Step4: Solve $f'(c) = 0$

Set $f'(c)=\frac{1}{2}-\frac{1}{2\sqrt{c}} = 0$. First, add $\frac{1}{2\sqrt{c}}$ to both sides: $\frac{1}{2\sqrt{c}}=\frac{1}{2}$. Then cross - multiply to get $2 = 2\sqrt{c}$. Divide both sides by 2: $\sqrt{c}=1$. Square both sides: $c = 1$.

Answer:

$c = 1$