Question

The value of \(c\) of Lagrange's Mean Value Theorem for \[ f(x)=\sqrt{25-x^2} \] on \([1,5]\) is:

• A. \(\sqrt{15}\)
• B. \(5\)
• C. \(\sqrt{10}\)
• D. \(1\)

Hint:

For Lagrange's Mean Value Theorem, always use: \[ f'(c)=\frac{f(b)-f(a)}{b-a} \] Then solve for \(c\) inside the open interval \((a,b)\).

Answer 1

The Correct Option is A

Concept:
According to Lagrange's Mean Value Theorem, if a function \(f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists at least one point \(c\in(a,b)\) such that: \(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}\)

Step 1: Write the given function and interval.
Here, \(\displaystyle f(x)=\sqrt{25-x^2}\) and the interval is: \(\displaystyle [1,5]\) So, \(\displaystyle a=1,\quad b=5\)

Step 2: Find \(f(1)\) and \(f(5)\).
\(\displaystyle f(1)=\sqrt{25-1^2}\) \(\displaystyle f(1)=\sqrt{24}=2\sqrt{6}\) Also, \(\displaystyle f(5)=\sqrt{25-5^2}\) \(\displaystyle f(5)=\sqrt{0}=0\)

Step 3: Find the average rate of change.
\(\displaystyle \frac{f(5)-f(1)}{5-1}=\frac{0-2\sqrt{6}}{4}\) \(\displaystyle =-\frac{\sqrt{6}}{2}\)

Step 4: Differentiate the function.
\(\displaystyle f(x)=\sqrt{25-x^2}=(25-x^2)^{1/2}\) \(\displaystyle f'(x)=\frac{1}{2}(25-x^2)^{-1/2}(-2x)\) \(\displaystyle f'(x)=\frac{-x}{\sqrt{25-x^2}}\)

Step 5: Apply Lagrange's Mean Value Theorem.
\(\displaystyle f'(c)=-\frac{\sqrt{6}}{2}\) So, \(\displaystyle \frac{-c}{\sqrt{25-c^2}}=-\frac{\sqrt{6}}{2}\) Removing the negative sign, \(\displaystyle \frac{c}{\sqrt{25-c^2}}=\frac{\sqrt{6}}{2}\)

Step 6: Square both sides.
\(\displaystyle \frac{c^2}{25-c^2}=\frac{6}{4}\) \(\displaystyle \frac{c^2}{25-c^2}=\frac{3}{2}\) Cross multiply: \(\displaystyle 2c^2=3(25-c^2)\) \(\displaystyle 2c^2=75-3c^2\) \(\displaystyle 5c^2=75\) \(\displaystyle c^2=15\) \(\displaystyle c=\sqrt{15}\) Hence, \(\displaystyle \boxed{c=\sqrt{15}}\)