Question

The function $f\left(x\right)=2x^3–3x^2–12x+4$, has

• A. two points of local maximum
• B. two points of local minimum
• C. one maxima and one minima
• D. no maxima or minima

Answer 1

The correct option is C one maxima and one minima

$\mathrm{We}\mathrm{have},f\left(x\right)=2x^3–3x^2–12x+4\therefore f\text{'}\left(x\right)=6x^2-6x-12\Rightarrow f\text{'}\left(x\right)=6\left(x^2-x-2\right)=6\left(x+1\right)\left(x-2\right)f\text{'}\left(x\right)=0\Rightarrow x=-1\mathrm{or}2.\therefore x=-1\mathrm{or}2\mathrm{are}\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{and}\mathrm{can}\mathrm{be}\mathrm{represented}\mathrm{as}$

From the above figure, x = -1 is a point of local maxima and x = 2 is a point of local minima.
So, f(x) has one minima and one maxima.