- x < 2
- x > 2
- x > 3
- 1 < x < 2
Solution:
f(x) = 2x3 - 9x2 + 12x + 29
⇒ f'(x) = 6x2 - 18x + 12
⇒ f'(x) = 6(x2 - 3x + 2)
⇒ f'(x) = 6(x - 1)(x - 2)
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ 6(x - 1)(x - 2) < 0
⇒ (x - 1)(x - 2) < 0
[Since, 6 > 0, 6(x - 1)(x - 2) < 0 ⇒ (x - 1)(x - 2) < 0]
⇒ 1 < x < 2
So, f(x) is decreasing for 1 < x < 2.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
($A$) The function $f$ is discontinuous exactly at one point in $(0,1)$
($B$) There is exactly one point in $(0,1)$ at which the function $f$ is continuous but $NOT$ differentiable
($C$) The function $\mathrm{f}$ is $NOT$ differentiable at more than three points in $(0,1)$
($D$) The minimum value of the function $f$ is $-\frac{1}{512}$