Find the c guaranteed by the Mean Value Theorem for the following function on the interval $[0,3]$.
\[
f(x)=x^{2}-1
\]
a. $c=9 / 2$
b. $c=2 / 9$
c. $c=3$
d. $c=3 / 2$

Step 1 ：Given the function \(f(x) = x^{2} - 1\) on the interval \([0,3]\), we need to find the value of \(c\) guaranteed by the Mean Value Theorem.

Step 2 ：The Mean Value Theorem states that if a function \(f\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a number \(c\) in the interval \((a, b)\) such that the derivative of the function at \(c\) is equal to the average rate of change of the function over the interval \([a, b]\).

Step 3 ：The derivative of the function \(f(x) = x^{2} - 1\) is \(f'(x) = 2x\).

Step 4 ：The average rate of change of the function over the interval \([0, 3]\) is \(\frac{f(3) - f(0)}{3 - 0}\).

Step 5 ：Substituting the values of \(f(3)\) and \(f(0)\) into the formula, we get \(\frac{8 - (-1)}{3 - 0} = 3\).

Step 6 ：We need to find the value of \(c\) such that \(f'(c) = \frac{f(3) - f(0)}{3 - 0}\).

Step 7 ：Solving the equation \(2c = 3\), we get \(c = \frac{3}{2}\).

Step 8 ：Final Answer: The value of \(c\) guaranteed by the Mean Value Theorem for the function \(f(x) = x^{2} - 1\) on the interval \([0,3]\) is \(\boxed{c = \frac{3}{2}}\).