Question:
The function $f(x) = x^2 \; e^{-2} x, x > 0$. Then the maximum value of $f(x)$ is
The function $f(x) = x^2 \; e^{-2} x, x > 0$. Then the maximum value of $f(x)$ is
Updated On: Jul 29, 2022
- $\frac{1}{e}$
- $\frac{1}{ 2 e}$
- $\frac{1}{e^2}$
- $\frac{4}{e^4}$
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The Correct Option is C
Solution and Explanation
Given : $f (x) = x^2 \; e^{-2x}, x > 0$
$f '(x) = x^2.e^{-2x} (-2) + e^{-2x} .2x$
put $f '(x) = 0 \; \Rightarrow \; 2e^{-2x}. x (-x + 1) = 0$
$\Rightarrow $ x = 1 or x = 0
$f"(x) = (-4x^2 - 6x + 1)e^{-2x}$
$f"(1) = -9e^{-2x} < 0$
$f"(0) = e^{-2x} > 0$
$\therefore$ value of f(x) is maximum at x = 1
$\because f\left(x\right)=x^{2}.e^{-2x}\Rightarrow f\left(1\right) =e^{-2} = \frac{1}{e^{2}} $
$f '(x) = x^2.e^{-2x} (-2) + e^{-2x} .2x$
put $f '(x) = 0 \; \Rightarrow \; 2e^{-2x}. x (-x + 1) = 0$
$\Rightarrow $ x = 1 or x = 0
$f"(x) = (-4x^2 - 6x + 1)e^{-2x}$
$f"(1) = -9e^{-2x} < 0$
$f"(0) = e^{-2x} > 0$
$\therefore$ value of f(x) is maximum at x = 1
$\because f\left(x\right)=x^{2}.e^{-2x}\Rightarrow f\left(1\right) =e^{-2} = \frac{1}{e^{2}} $
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
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This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
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- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
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