CUET UG 2025 — Mathematics Calculus

To find the local minima of $f(x) = 4x^3 - 7x^2$f(x)=4x3−7x2, find where the derivative equals zero.

The first derivative is:

$$f'(x) = 12x^2 - 14x$$f′(x)=12x2−14x

Setting the derivative equal to zero:

$$12x^2 - 14x = 0$$12x2−14x=0

$$2x(6x - 7) = 0$$2x(6x−7)=0

This gives two critical points:

$x = 0$x=0 or $x = \frac{7}{6}$x=67​

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To determine which critical points are local minima, use the second derivative test.

The second derivative is:

$$f''(x) = 24x - 14$$f′′(x)=24x−14

At a critical point:

• If $f''(x) > 0$f′′(x)>0, the point is a local minimum
• If $f''(x) < 0$f′′(x)<0, the point is a local maximum

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For $x = 0$x=0:

$$f''(0) = 24(0) - 14$$f′′(0)=24(0)−14

$$f''(0) = -14 < 0$$f′′(0)=−14<0

This is a local maximum.

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For $x = \frac{7}{6}$x=67​:

$$f''\left(\frac{7}{6}\right) = 24 \times \frac{7}{6} - 14$$f′′(67​)=24×67​−14

$$f''\left(\frac{7}{6}\right) = 28 - 14$$f′′(67​)=28−14

$$f''\left(\frac{7}{6}\right) = 14 > 0$$f′′(67​)=14>0

This is a local minimum.

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Therefore, the function has a point of local minima at $x = \frac{7}{6}$x=67​.