Question

examine the function f(x)= x^3-9x^2+24x for maxima and minima

Answer 1

it should be your answer.

Answer 2

Hey there!

Finding the maximum and minimum points of a function requires a first/second derivative function for finding the critical points;

If we presume that the given variable of x = c is a specific critical point for a function of variable "x" then the following condition applies;

If f'(x) > 0 to it's left side of x = c and f'(x) < 0 to the right side of x = c then x = c will be a local maximum.

If f'(x) < 0 to it's left side of x = c and f'(x) > 0 to it's right side of x = c then x = c will be a local minimum.

And,  If f'(x) is having the similar sign on both of it's sides of the critical point x = c then x = c is said to've no local maximum or minimum points.

A Critical point of a particular given function is defined as the points in which the function is defined whereas the derivative is either zero or undefined.

$\bf{\therefore \quad \dfrac{d}{dx} (x^3 - 9x^2 + 24x)}$

$\bf{\therefore \quad \dfrac{d}{dx} (x^3) - \dfrac{d}{dx} (9x^2) + \dfrac{d}{dx} (24x)}$

$\bf{\therefore \quad 3x^{3 - 1} - 9 \dfrac{d}{dx} (x^2) + 24 \dfrac{d}{dx} (x)}$

$\bf{\therefore \quad 3x^2 - 9 \times 2x^{2 - 1} + 24 \times 1}$

$\bf{\therefore \quad 3x^2 - 18x + 24 = 0}$

Solving this equation via quadratic formula;

For a given quadratic equation of the given form of a equation of $\bf{ax^2 + bx + c = 0}$ for which the solutions are defined as;

$\boxed{\bf{x_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}}$

Here, a = 3,  b = - 18  and  c = 24.\

$\bf{x_1 = \dfrac{- (- 18) + \sqrt{(- 18)^2 - 4 \times 3 \times 24}}{2 \times 3}}$

$\bf{x_1 = \dfrac{18 + \sqrt{324 - 288}}{6}}$

$\bf{x_1 = \dfrac{18 + \sqrt{36}}{6}}$

$\bf{x_1 = \dfrac{18 + 6}{6}}$

$\bf{x_1 = \dfrac{24}{6}}$

$\boxed{\bf{\underline{x_1 = 4}}}$

For the negative value,

$\bf{x_2 = \dfrac{- (- 18) - \sqrt{(- 18)^2 - 4 \times 3 \times 24}}{2 \times 3}}$

$\bf{x_2 = \dfrac{18 - \sqrt{324 - 288}}{6}}$

$\bf{x_2 = \dfrac{18 - \sqrt{36}}{6}}$

$\bf{x_2 = \dfrac{18 - 6}{6}}$

$\bf{x_2 = \dfrac{12}{6}}$

$\boxed{\bf{\underline{x_2 = 2}}}$

Find the domain of this function; The domain of the function is defined as per the set of the inputs or the "argument values" for which the functions are becoming "real" and "defined".

This function has no particular definite points nor does it've domain constraints. Therefore the domain of this function is in the set of;

$\boxed{\bf{- \infty < x < \infty}}$

Hence, all the following critical points of, x =2 and x = 4,  are in the function's domain.

Now, join the critical points and the function's domain to get the monotone intervals of the function that is;

$\bf{- \infty < x < 2, \: \: 2 < x < 4, \: \: 4 < x < \infty}$

Now, check the following signs for the function of,  f'(x) = 3x^2 - 18x + 24 for each and every monotone interval of the given function.

1) Check the sign for 3x^2 - 18x + 24 at the monotone interval of $\bf{- \infty < x < 2}$.

Substitute the following value by evaluation of derivative at a point of that particular monotone interval, (in the interval), x = 1.

$\bf{3 \times 1 ^2 - 18 \times 1 + 24}$

$\bf{9}$

Which is the sign of positive with an increasing value (behavior).

2) Check the sign for 3x^2 - 18x + 24 at the monotone interval of 2 < x < 4.

Substitute the following value by evaluation of derivative at a point of that particular monotone interval, (in the interval), x = 3.

$\bf{3 \times 3^2 - 18 \times 3 + 24}$

$\bf{- 3}$

Which is the sign of negative with a decreasing value (behavior).

3) Check the sign for 3x^2 - 18x + 24 at the monotone interval of $\bf{4 < x < \infty}$.

Substitute the following value by evaluation of derivative at a point of that particular monotone interval, (in the interval), x = 5.

$\bf{3 \times 5^2 - 18 \times 5 + 24}$

$\bf{9}$

Which is the sign of positive with an increasing value (behavior).

[CHECK THE ATTACHMENTS FOR THE FOLLOWING MONOTONE INTERVAL BEHAVIOR OF INCREASEMENT OR DECREASEMENT OF VALUES WHEN INCORPORATING THE SIGN OF THE MONOTONE INTERVAL]

Insert the extreme points for the given interval values of "x = 2" into the equation of "x^3 - 9x^2 + 24x":

$\bf{(2)^3 - 9(2)^2 + 24(2)}$

$\bf{8 - 36 + 48}$

$\bf{- 28 + 48}$

$\boxed{\bf{\underline{\therefore \quad Maximum \: \: (2, \: 20)}}}$

Insert the extreme points for the given interval values of "x = 4" into the equation of "x^3 - 9x^2 + 24x":

$\bf{(4)^3 - 9(4)^2 + 24(4)}$

$\bf{64 - 144 + 96}$

$\bf{- 80 + 96}$

$\boxed{\bf{\underline{\therefore \quad Minimum \: \: (4, \: 16)}}}$

FINAL ANSWER:

MAXIMUM  (2, 20);   MINIMUM  (4, 16).

Which is the required fully detailed solution for these types of queries.

Hope this helps you and clears your doubts for finding the maxima and minima for a particular function of the equation !!!