Question

How do you find the maxima and minima of the function $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-24x+3$ ?

Answer 1

Hint: To start with, we have, a function given as, $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-24x+3$ and we are to find the maxima and minima of the problem here. So, we can start it with finding the first order and second order derivatives and equaling the first order derivative with zero will give us the points of maxima and minima. Then putting the values into the second order derivative, we will get on which point we have the maximum or minima.

Complete step-by-step solution:
According to the problem, we are given, $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-24x+3$
Now, to find the maxima and minima, we are to find the first order derivative of our given function.
So, the first order derivative of the function, $f'\left( x \right)$ would be, $f'\left( x \right)=3{{x}^{2}}+6x-24$
Again, at any point of maxima and minima we get, $f'\left( x \right)=0$.
Starting with, $f'\left( x \right)=3{{x}^{2}}+6x-24=0$
Dividing by 3 on both sides,
${{x}^{2}}+2x-8=0$
Now, we can divide 8 into 4 and 2 and simplify it with the middle term factor.
${{x}^{2}}+4x-2x-8=0$
Taking x common from first two terms, and -2 from the last two, we get, $x\left( x+4 \right)-2\left( x+4 \right)=0$
Simplifying, $\left( x-2 \right)\left( x+4 \right)=0$
Thus, we are getting the values of x as, x = 2, - 4.
Putting the value of x = 2, we have, $f\left( 2 \right)={{2}^{3}}+3\times {{2}^{2}}-24\times 2+3$
Thus, we get the value as, $f\left( 2 \right)=8+12-48+3=-25$ which gives us a negative value.
Again, Putting the value of x = -4, we have, $f\left( -4 \right)={{\left( -4 \right)}^{3}}+3\times {{\left( -4 \right)}^{2}}-24\times \left( -4 \right)+3$
Thus, we get the value as, $f\left( -4 \right)=-64+48+96+3=83$ which gives us a positive value.
Again, for the second order derivative, $f''\left( x \right)=6x+6$
When x = - 4, we get, $f''\left( x \right)=6\times \left( -4 \right)+6=-24+6=-18$, so we will get the maximum value in -4.
When x = 2, we get, $f''\left( x \right)=6\times 2+6=12+6=18$, so we will get the minimum value in 2.
Then, we get maxima at -4 and minima at point 2.

Note: The maxima of a function f(x) are all the points on the graph of the function which are 'local maximums'. A point where x = a is a local maximum if, when we move a small amount to the left (points with x < a) or right (points with x > a), the value of f(x) decreases. We can visualize this as our graph having the peak of a 'hill' at x=a. Similarly, the minima of f(x) are the points for which, when we move a small amount to the left or right, the value of f(x) increases. We call these points 'local minimums', and we can visualize them as the bottom of a 'trough' in our graph.