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:

Logic Tip: Always double-check that your final calculated value for $c$ strictly falls within the open interval $(a,b)$. If you get multiple roots, only those satisfying $a<c<b$ are valid answers under the theorem.

Answer 1

The Correct Option is A

Concept:
Lagrange's Mean Value Theorem (LMVT) states that if a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one value $c \in (a, b)$ such that: $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$
Step 1: Evaluate the function at the endpoints of the interval.
Here, $f(x) = \sqrt{25 - x^2}$, and the interval is $[1, 5]$. $$ f(1) = \sqrt{25 - 1} = \sqrt{24} $$ $$ f(5) = \sqrt{25 - 25} = 0 $$
Step 2: Calculate the derivative of the function.
$$ f'(x) = \frac{d}{dx}(\sqrt{25 - x^2}) = \frac{1}{2\sqrt{25 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{25 - x^2}} $$ So, $$ f'(c) = \frac{-c}{\sqrt{25 - c^2}} $$
Step 3: Apply LMVT and solve for $c$.
$$ \frac{-c}{\sqrt{25 - c^2}} = \frac{f(5) - f(1)}{5 - 1} $$ $$ = \frac{0 - \sqrt{24}}{4} = -\frac{\sqrt{24}}{4} $$ Squaring both sides: $$ \frac{c^2}{25 - c^2} = \frac{24}{16} = \frac{3}{2} $$ Cross-multiplying: $$ 2c^2 = 3(25 - c^2) $$ $$ 2c^2 = 75 - 3c^2 $$ $$ 5c^2 = 75 \Rightarrow c^2 = 15 $$ $$ c = \pm \sqrt{15} $$ Since $c \in (1, 5)$: $$ \boxed{c = \sqrt{15}} $$