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closed as a duplicate of: GATE CSE 2023 | Question: 18
Let $f(x)=x^3+15 x^2-33 x-36$ be a real valued function. Which statement is/are TRUE?
1. $f(x)$ has a local maximum.
2. $f(x)$ does NOT have a local maximum.
3. $f(x)$ has a local minimum.
4. $f(x)$ does NOT have a local minimum.

the real valued function $f(x)=x^3+15x^2-33x-36=0$

1) find $f’(x)=0$

$\implies3x^2+30x-33=0$

$\implies x^2+10x-11=0$

$\implies x^2+11x-x-11=0$

$\implies (x+11)(x-1)=0$

$\implies x=-11,1$

2) find $f^”(x)$ we get : $f^”(x)=6x+30$

​​​​​​​3 )  $f^”(1)=6+30=36>0$ it is gives local minima

4  $f^”(-11)=-66+30=-36<0$ it is gives local maxima.

so given function $f(x)$ will give local maxima at $x=-11$ and local minima at $x=1$

$\therefore$ Option $A,C$ is correct.
There can be two ways to solve this question.

$f(x) = x^3 +15x^2 -33x-36$

$\textbf{Method 1 (By Observation):}$

$f'(x) = 3x^2 + 30x - 33 =3(x^2 + 10x -11)$

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 $(-5,\infty).$

$\textbf{Therefore (A),(C)}$

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 and based on that we can say one local maxima and minima exist for the original graph $f(x).$

$\textbf{Method 2 (By General Procedure ):}$

$f'(x) = 3 \left((x+5)^2- 36 \right)$

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 $x=1$)

$\textbf{Therefore (A),(C)}$

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