Question:
The value of $c$ from the Lagrange�s mean value theorem for which $f(x) = \sqrt{25 - x^2}$ in $[1,5]$, is
The value of $c$ from the Lagrange�s mean value theorem for which $f(x) = \sqrt{25 - x^2}$ in $[1,5]$, is
Updated On: Apr 19, 2024
- $5$
- $1$
- $\sqrt{15}$
- None of these
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The Correct Option is C
Solution and Explanation
It is clear that $f (x)$ has a definite and unique value for each $x \in [1, 5].$
Thus, for every point in the interval $[1, 5],$ the value of $f (x)$ exists.
So, $f(x)$ is continuous in the interval [1, 5].
Also, $f(x) = \frac{ - x }{\sqrt{ 25 - x^2}}$, which clearly exists for all x in an open interval (1, 5).
Hence, $f'(x)$ is differentiable in $(1,5)$.
So, there must be a value $c \in [1, 5]$ such that
$f'(c) = \frac{f(5) - f(1) }{5 -1} = \frac{ 0 - \sqrt{24}}{4}$
$= \frac{ 0 - 2 \sqrt{6}}{4} = \frac{- \sqrt{6}}{2}$
But $f'(c) = \frac{ - c}{\sqrt{25 - c^2}}$
$\therefore \frac{-c}{\sqrt{25 - c^{2}}} = - \frac{\sqrt{6}}{2} $
$\Rightarrow 4 c^{2} = 6 \left(25 - c^{2}\right)$
$\Rightarrow 4c^{2} = 150 - 6c^{2} \Rightarrow 10c^{2} = 150 $
$\Rightarrow c^{2 } = 15 \Rightarrow c = \pm\sqrt{15} $
$\therefore c = \sqrt{15} \in\left[1,5\right] $
Thus, for every point in the interval $[1, 5],$ the value of $f (x)$ exists.
So, $f(x)$ is continuous in the interval [1, 5].
Also, $f(x) = \frac{ - x }{\sqrt{ 25 - x^2}}$, which clearly exists for all x in an open interval (1, 5).
Hence, $f'(x)$ is differentiable in $(1,5)$.
So, there must be a value $c \in [1, 5]$ such that
$f'(c) = \frac{f(5) - f(1) }{5 -1} = \frac{ 0 - \sqrt{24}}{4}$
$= \frac{ 0 - 2 \sqrt{6}}{4} = \frac{- \sqrt{6}}{2}$
But $f'(c) = \frac{ - c}{\sqrt{25 - c^2}}$
$\therefore \frac{-c}{\sqrt{25 - c^{2}}} = - \frac{\sqrt{6}}{2} $
$\Rightarrow 4 c^{2} = 6 \left(25 - c^{2}\right)$
$\Rightarrow 4c^{2} = 150 - 6c^{2} \Rightarrow 10c^{2} = 150 $
$\Rightarrow c^{2 } = 15 \Rightarrow c = \pm\sqrt{15} $
$\therefore c = \sqrt{15} \in\left[1,5\right] $
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Concepts Used:
Mean Value Theorem
The theorem states that for a curve f(x) passing through two given points (a, f(a)), (b, f(b)), there is at least one point (c, f(c)) on the curve where the tangent is parallel to the secant passing via the two given points is the mean value theorem.
The mean value theorem is derived herein calculus for a function f(x): [a, b] → R, such that the function is continuous and differentiable across an interval.
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- The function f(x) = differentiable across the interval (a, b).
- A point c exists in (a, b) such that f'(c) = [ f(b) - f(a) ] / (b - a)