Find the intervals on which the function $f (x) = 4 x^{3} - 9 x^{2} + 3 x + 5$ is increasing.

easy 2 marks Class 12 · Mathematics · Application of Derivatives

Answer

(- \infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)

The correct answer is $(- \infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$ — Thus $f$ is increasing on $(- \infty, \frac{1}{2})$ and $(\frac{3}{2}, \infty)$ , but decreasing on $(\frac{1}{2}, \frac{3}{2})$ .

Step-by-step solution

Differentiate $f (x)$ to get $f^{'} (x) = 12 x^{2} - 18 x + 3$ .
Factor $f^{'} (x)$ as $3 (4 x^{2} - 6 x + 1) = 3 (2 x - 1) (2 x - 3)$ .
Set each factor to zero to find critical points: $2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$ , $2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$ .
Determine sign of $f^{'} (x)$ on intervals $(- \infty, \frac{1}{2})$ , $(\frac{1}{2}, \frac{3}{2})$ , $(\frac{3}{2}, \infty)$ .
On $(- \infty, \frac{1}{2})$ and $(\frac{3}{2}, \infty)$ , $f^{'} (x) > 0$ ; on $(\frac{1}{2}, \frac{3}{2})$ , $f^{'} (x) < 0$ .
Thus $f$ is increasing on $(- \infty, \frac{1}{2})$ and $(\frac{3}{2}, \infty)$ , but decreasing on $(\frac{1}{2}, \frac{3}{2})$ .

Common mistake: Not factoring the derivative correctly, leading to wrong critical points

Concept tested: Increasing/decreasing intervals

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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Find the intervals on which the function f(x)=4x3−9x2+3x+5 is increasing.

Step-by-step solution

Find the intervals on which the function $f (x) = 4 x^{3} - 9 x^{2} + 3 x + 5$ is increasing.