JFI. Plot of the provided function.

26 votes

A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve $3x^4-16x^3+24x^2+37$ is

- $0$
- $1$
- $2$
- $3$

35 votes

Best answer

Answer is $(b)$.

$f(x)=3x^4-16x^3+24x^2+37$

$f'(x)=12x^3-48x^2+48x=0$

$\implies 12x\left(x^2-4x+4\right)=0$

$x\left(x-2\right)^2=0 $

$\implies x=0, 2$

$f''(x)=36x^2-96x+48$

at $x=0, f''(x)=48>0$ it means that $x =0$ is local minima.

but at $x=2, f''(x)=0$ so we can't apply second derivative test. So, we can apply first derivative test.

$f'(1) = 12, f'(3) = 36 $. So, $f'(x)$ is not changing sign on either side of 2. So, $x=2$ is neither maxima nor minima.

So, only one extremum i.e. x=0.