Find the location(s) and value(s) of all absolute extrema (state which) for f(x)= x^3 - 6x^2 + 9x...


Find the location(s) and value(s) of all absolute extrema (state which) for {eq}f(x)= x^3 - 6x^2 + 9x - 8 {/eq} on the domain [2,5].

Critical Points:

The critical points are the points which belong to the domain of the function and at these points, the function is defined but the derivative of the function is undefined or zero.


Any point {eq}p {/eq} is said to be a critical point of a function {eq}f(x) {/eq} if either

1. {eq}f^{\prime}(p)=0 {/eq},


2. {eq}f^{\prime}(p) {/eq} doesn't exist.

Answer and Explanation:

The function is given as

{eq}\displaystyle f(x)= x^3 - 6x^2 + 9x - 8 {/eq}

and we want to find the exact location and value of absolute extrema of the function on the interval {eq}\displaystyle [2,5] {/eq}.

First, to find the critical points of the function we will use the derivative test.

{eq}\displaystyle \begin{align} &\therefore f^{\prime}(x) = 0\\[0.3cm] &\Rightarrow \frac{d}{dx}f(x) = 0\\[0.3cm] &\Rightarrow \frac{d}{dx}\left(x^3-6x^2+9x-8\right)=0\\[0.3cm] &\Rightarrow \frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(9x\right)-\frac{d}{dx}\left(8\right)=0\\[0.3cm] &\Rightarrow 3x^2-12x+9=0\\[0.3cm] &\Rightarrow 3(x^2-4x+3)=0\\[0.3cm] &\Rightarrow x^2-4x+3=0\\[0.3cm] &\Rightarrow x^2-3x-x+3=0\\[0.3cm] &\Rightarrow x(x-3)-1(x-3)=0\\[0.3cm] &\Rightarrow (x-3)(x-1)=0\\[0.3cm] &\Rightarrow x=3,~x=1.\\[0.3cm] \end{align} {/eq}

Therefore the critical points are {eq}\displaystyle x=1,~x=3 {/eq}

but the point {eq}\displaystyle x=1 \notin [2,5] {/eq}

so the only critical point in the given domain is {eq}\displaystyle x=3 {/eq}.

Now for the absolute extreme values of the function, we will evaluate the function at the endpoints of the interval as well as at the critical points.

{eq}\displaystyle \begin{align} &f(2) = (2)^3 - 6(2)^2 + 9(2) - 8 = 8-24+18-8 = -6\\[0.3cm] &f(3) = (3)^3 - 6(3)^2 + 9(3) - 8 = 27-54+27-8 = -8\\[0.3cm] &f(5) = (5)^3 - 6(5)^2 + 9(5) - 8 = 125-150+45-8 = 12\\[0.3cm] \end{align} {/eq}

From the above calculations, the given function has an absolute maximum at the points {eq}\displaystyle x=5 {/eq}

and the given function has an absolute minimum at the points {eq}\displaystyle x=3 {/eq}.

Learn more about this topic:

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9

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