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

Answer: e^(1/e). Solution: To find the maximum value of f(x) = (1/x)^x, we use calculus to find where the derivative equals zero. Rewrite the function in a simpler form: f(x) = (1/x)^x f(x) = x^(-x) Taking the natural log of both sides: f(x) = (x^(-x)) f(x) = -x x Differentiating both sides: (f'(x))/(f(x)) = - x - 1 f'(x) = f(x)(- x - 1) f'(x) = x^(-x)(- x - 1) For critical points, set f'(x) = 0: x^(-x)(- x - 1) = 0 Since x^(-x) is always positive for x > 0: - x - 1 = 0 x = -1 x = e^(-1) x = 1/e Substituting x = 1/e into the original function: f(1/e) = (1/(1/e))^(1/e) f(1/e) = e^(1/e) Therefore, the maximum value of f(x) = (1/x)^x is e^(1/e).

Calculus

Application of Derivatives

Medium

core

The maximum value of f(x) = $\left(\frac{1}{x}\right)^x$ is

$e^{-1/e}$

$\left(\frac{1}{e}\right)^e$

$e^{1/e}$

$\left(\frac{1}{e}\right)^{-e}$

✅ Correct Option: 3

To find the maximum value of $f(x) = \left(\frac{1}{x}\right)^x$ , we use calculus to find where the derivative equals zero.

Rewrite the function in a simpler form:

$f(x) = \left(\frac{1}{x}\right)^x$

$f(x) = x^{-x}$

Taking the natural log of both sides:

$\ln f(x) = \ln(x^{-x})$

$\ln f(x) = -x \ln x$

Differentiating both sides:

$\frac{f'(x)}{f(x)} = -\ln x - 1$

$f'(x) = f(x)(-\ln x - 1)$

$f'(x) = x^{-x}(-\ln x - 1)$

For critical points, set $f'(x) = 0$ :

$x^{-x}(-\ln x - 1) = 0$

Since $x^{-x}$ is always positive for $x > 0$ :

$-\ln x - 1 = 0$

$\ln x = -1$

$x = e^{-1}$

$x = \frac{1}{e}$

Substituting $x = \frac{1}{e}$ into the original function:

$f\left(\frac{1}{e}\right) = \left(\frac{1}{1/e}\right)^{1/e}$

$f\left(\frac{1}{e}\right) = e^{1/e}$

Therefore, the maximum value of $f(x) = \left(\frac{1}{x}\right)^x$ is $e^{1/e}$ .

2025: 13 May Shift 2

The sum of two positive numbers is 60. If the sum of their squares in minimum, then the absolute value of the difference of their cubes is

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. Maximum value of $f(x) = -\lvert x+1 \rvert + 3$	I. 6
B. Minimum value of $f(x) = (2x-1)^2 + 5$	II. 5
C. Maximum value of $f(x) = 6 - x^2$	III. no maximum value
D. Maximum value of $f(x) = x^3 + 1$	IV. 3

Choose the correct answer from the options given below:

2022: 30 Aug Shift 1

The given function $f(x) = x^5 - 5x^4 + 5x^3 - 1$ ; has/have

(a) local maxima at $x = 1$

(b) local maximum value is 0

(d) local minimum value is $-28$

(e) The point of inflexion is $x = 1$

Choose the correct answer from the options given below

2022: 4 Aug Shift 1

The interval in which the function $f(x) = 2x^3 + 3x^2 - 12x + 1$ is strictly increasing is -

2026: 25 May Shift 1

Which of the following statements are correct ?

A. The function $f(x) = a^x$ is increasing on $(-\infty,\infty)$ if $a>1$ .

B. $f(x) = x^2 - 2x$ is decreasing in $(-\infty,1)$ .

C. $f(x) = x(x-3)^2$ increases for $1 \le x \le 3$ .

D. $f(x) = e^{-x}$ decreases in $[0,\infty)$ .

Choose the correct answer from the options given below:

2022: 7 Aug Shift 2

If $2x + y = 6$ then the maximum value of $x^2 y$ is :

The maximum value of f(x) = (1x)x\left(\frac{1}{x}\right)^x(x1​)x is

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The maximum value of f(x) = $\left(\frac{1}{x}\right)^x$ is