# GATE2008-25

3.3k views

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

1. $0$
2. $1$
3. $2$
4. $3$
in Calculus
edited
9

JFI. Plot of the provided function. 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.

edited
17
When we can't tell about maxima/minima using second derivative test then we can also take help of third derivative test.

iff $f'''(x)\neq 0$ then there won't be any maxima/minima

iff $f'''(x)= 0$ then check for $f''''(x)$ if $+ve$ then minima , if $-ve$ then maxima.
3
Why we cant apply second derivative test? can someone please explain... when is it possible to apply second derivative test?

Also, why don't we go for third dera'ivative test?

## Related questions

1
4k views
$\lim_{x \to \infty}\frac{x-\sin x}{x+\cos x}$ equals $1$ $-1$ $\infty$ $-\infty$
Find the minimum value of $3-4x+2x^2$.
If $f(x)$ is defined as follows, what is the minimum value of $f(x)$ for $x \in (0, 2]$ ? $f(x) = \begin{cases} \frac{25}{8x} \text{ when } x \leq \frac{3}{2} \\ x+ \frac{1}{x} \text { otherwise}\end{cases}$ $2$ $2 \frac{1}{12}$ $2\frac{1}{6}$ $2\frac{1}{2}$