The maximum value of x^(-x) is Option 1: e^(e-1) Option 2: e^(1/e) Option 3: e^((-1)/e) Option 4: e^((-4)/e)

Answer: e^(1/e). Solution: Let f(x) = x^(-x). Taking logarithm: f(x) = -x x. Differentiating: (f'(x))/(f(x)) = -(1+ x). Setting f'(x) = 0: 1 + x = 0 x = 1/e. Maximum value = (1/e)^(-1/e) = e^(1/e).

Calculus

Application of Derivatives

Medium

The maximum value of $x^{-x}$ is

$e^{e-1}$

$e^{\frac{1}{e}}$

$e^{\frac{-1}{e}}$

$e^{\frac{-4}{e}}$

✅ Correct Option: 2

Let $f(x) = x^{-x}$ . Taking logarithm: $\ln f(x) = -x \ln x$ . Differentiating: $\frac{f'(x)}{f(x)} = -(1+\ln x)$ . Setting $f'(x) = 0$ : $1 + \ln x = 0 \Rightarrow x = 1/e$ . Maximum value $= (1/e)^{-1/e} = e^{1/e}$ .

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The function, $f(x) = x - \frac{1}{x}$ is

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For the function $f(x) = e^{-2x}(2-x)^2$ , the point of local maxima is:

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If an edge of a variable cube is increasing at the rate of 0.8 cm/sec, at what rate its surface area is increasing when its edge is 10 cm?

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The side of an equilateral triangle is increasing at the rate of 2 cm/sec. The rate at which the area of the triangle is increasing, when the side of the triangle is 5 cm, is

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The revenue function is given by $R(x)=150x-2x^2-x^3$ , where $x$ is the number of units demanded, then the marginal revenue when 5 units are sold, is :

The maximum value of x−xx^{-x}x−x is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The maximum value of $x^{-x}$ is