CUET UG 2025 — Mathematics Calculus

The function $f(x) = x + \frac{1}{x}$f(x)=x+x1​ has a local maximum or minimum where the derivative equals zero.

$$f'(x) = 1 - \frac{1}{x^2}$$f′(x)=1−x21​

Setting the derivative equal to zero:

$$1 - \frac{1}{x^2} = 0$$1−x21​=0

$$1 = \frac{1}{x^2}$$1=x21​

$$x^2 = 1$$x2=1

$x = 1$x=1 or $x = -1$x=−1

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To determine whether each critical point is a maximum or minimum, use the second derivative test.

$$f''(x) = \frac{2}{x^3}$$f′′(x)=x32​

When $f''(x) < 0$f′′(x)<0, the function has a local maximum.
When $f''(x) > 0$f′′(x)>0, the function has a local minimum.

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At $x = -1$x=−1:

$$f''(-1) = \frac{2}{(-1)^3}$$f′′(−1)=(−1)32​

$$f''(-1) = \frac{2}{-1}$$f′′(−1)=−12​

$$f''(-1) = -2$$f′′(−1)=−2

Since $f''(-1) < 0$f′′(−1)<0, the function has a local maximum at $x = -1$x=−1.

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At $x = 1$x=1:

$$f''(1) = \frac{2}{(1)^3}$$f′′(1)=(1)32​

$$f''(1) = 2$$f′′(1)=2

Since $f''(1) > 0$f′′(1)>0, the function has a local minimum at $x = 1$x=1.

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Therefore, the function has a local maximum at $x = -1$x=−1 and a local minimum at $x = 1$x=1.