Method 0 (Formula Based):
1) A cubic function $ax^3+bx^2+cx+d$ does not have local maxima and local minima iff $b^2 \leq 3ac.$
2) A cubic function $ax^3+bx^2+cx+d$ have a local maxima and a local minima iff $b^2 > 3ac.$
The point for which local minima occurs is given by $x= \frac{-b + \sqrt{b^2-3ac}}{3a}$ and the point for local maxima occurs is given by $x= \frac{-b - \sqrt{b^2-3ac}}{3a}$
Formulae are similar to roots of a quadratic equation, so these can be remembered and if you can't remember then you can go with next two methods.
So, here $a=1,b=15,c=-33$ for the given cubic function $x^3 + 15x^2 -33x -36.$
Since, $b^2 >3ac$ i.e. $15^2 > 3 \times 1 \times (-33).$ So, given cubic function have a local maxima and a local minima. Hence, B and D are correct options.
Proof: Assuming $f: \mathbb{R} \rightarrow \mathbb{R}$
$f(x)=ax^3+bx^2+cx+d; a \neq 0$
$f'(x)= 3ax^2+2bx+c$
$f'(x)=0 \Rightarrow$ $3ax^2+2bx+c=0$ which gives:
$$x=\frac{-b\pm \sqrt{b^2-3ac}}{3a}$$
Now, $f''(x) = 6ax+2b = 2(3ax+b)$ and ignoring "2" from $f''(x)$ because it does not matter to determine the sign of $f''(x).$ So, consider, $f''(x)=3ax+b$
$f''(\frac{-b +\sqrt{b^2-3ac}}{3a})=\sqrt{b^2-3ac} >0$
$f''(\frac{-b -\sqrt{b^2-3ac}}{3a})=-\sqrt{b^2-3ac} <0$
Hence, the point for which local minima occurs is given by $x= \frac{-b + \sqrt{b^2-3ac}}{3a}$ and the point for local maxima occurs is given by $x= \frac{-b - \sqrt{b^2-3ac}}{3a}$
Here, note that $x$ would be real if $b^2-3ac\geq 0$ but when $b^2-3ac= 0$ then $f''(.)=0$ and so, when $b^2=3ac,$ $x$ would be an inflection point (neither maxima nor minima) which is given by $x= \frac{-b}{3a}.$ Hence, local maxima and local minima exist when $b^2 > 3ac.$
Method 1 (By Observation):
$f'(x) = 3x^2 + 30x - 33 =3(x^2 + 10x -11)$
Now, you can see that $f'(x)$ is an upward parabola because coefficient of $x^2$ is positive. You can also write $f'(x)$ as:
$f'(x) = 3 \left((x+5)^2- 36 \right) $
Now, Since, $(x+5)^2 \geq 0,$ So, you can see for $x=-5,$ $f'(x)$ has the lowest value which is $-3\times 36$ which is negative and since it is upward parabola, so it cuts the x-axis as well and goes up. Hence, $f'(x)$ goes positive to negative and then negative to positive on the x-axis (from left to right).
Hence,
$1)$ $f'(x)$ goes positive to negative for the interval $(-\infty, -5)$ and so, slope of a straight line goes positive to negative for points in the interval $(-\infty, -5)$ for the graph of $f(x).$ Therefore, $f(x)$ has one local maxima in the interval $(-\infty, -5)$ and
$2)$ $f'(x)$ goes negative to positive for the interval $(-5,\infty)$ and so, slope of a straight line goes negative to positive for points in the interval $(-5,\infty)$ for the graph of $f(x).$ Therefore, $f(x)$ has one local minima in the interval $(-5,\infty).$
$\textbf{Therefore (B),(D)}$
You can also plot $f'(x)$ easily as:
$(1)$ Just make the graph of $x^2$
$(2)$ Shift $5$ units left on the x-axis, it will give $(x+5)^2$
$(3)$ Shift the new graph i.e $(x+5)^2,$ $36$ units downward which will give $(x+5)^2 -36$
$(4)$ Multiplication of $3$ is working as a stretching factor means each value of $(x+5)^2 -36$ will be multiplied by $3$ but based on $(x+5)^2 -36,$ we can say that the graph of $f'(x)$ is upward parabola and it goes from positive to negative and then negative to positive (left to right) and based on that we can say that one local maxima and one local minima exist for the graph $f(x).$
Method 2 (By General Procedure):
$f'(x) = 3 \left((x+5)^2- 36 \right) $
Now, $f'(x)=0$ means $(x+5)^2-6^2=0$
$x=-11,1$
$f''(x)=6x+30$
$f''(-11)<0$ (local maxima at $x=-11$)
$f''(1)>0$ (local minima at $x=1$)
$\textbf{Therefore (B),(D)}$
