the mean value theorem guarantees the existence of a special point on the graph of y = √x between (0,0) and…

Explanation:

Step1: Recall Mean - Value Theorem

The Mean - Value Theorem for a function $y = f(x)$ on the interval $[a,b]$ states that if $y = f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in(a,b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b = 4$, $f(x)=\sqrt{x}$, $f(a)=f(0)=0$, $f(b)=f(4)=2$.

Step2: Calculate the derivative of $y = f(x)$

The derivative of $y=\sqrt{x}=x^{\frac{1}{2}}$ using the power - rule $(x^{n})^\prime=nx^{n - 1}$ is $y^\prime=f^\prime(x)=\frac{1}{2\sqrt{x}}$.

Step3: Calculate the slope of the secant line

The slope of the secant line connecting the points $(0,0)$ and $(4,2)$ is $\frac{f(4)-f(0)}{4 - 0}=\frac{2-0}{4}=\frac{1}{2}$.

Step4: Solve for $x$ (the $x$ - coordinate of the point)

Set $f^{\prime}(x)=\frac{1}{2}$, so $\frac{1}{2\sqrt{x}}=\frac{1}{2}$. Cross - multiply to get $2 = 2\sqrt{x}$, then $\sqrt{x}=1$, and $x = 1$.

Step5: Calculate the $y$ - coordinate of the point

Substitute $x = 1$ into $y=\sqrt{x}$, we get $y = 1$.

Answer:

B. $(1,1)$