# Find the absolute maximum and minimum of the given function on the specified interval...

## Question:

Find the absolute maximum and minimum of the given function on the specified interval {eq}f(t)=6t^5-40t^3, \ -3 \leq t \leq 0 {/eq}

## Absolute Maxima and Minima in an Interval

{eq}{/eq}

To find the absolute maxima and minima for a function y = f(x) in an interval from x = a to x = b (both inclusive) , we need to follow the steps below :

1. Find the values of f(x) at the corner points of the interval, i.e., determine f(a) and f(b).

2. Find the local maxima and minima (whichever exists) for f(x) in the given interval using derivatives.

3. Compare their values with f(a) and f(b) and arrange all the values obtained to get the absolute maxima and minima.

## Answer and Explanation:

{eq}{/eq}

Given function :

$$\displaystyle f(t)=6t^5-40t^3 \quad \quad -3 \leq t \leq 0 \\$$

The values of f(t) at the corner points of given interval are :

{eq}\displaystyle{f(-3) = 6(-3)^5 - 40(-3)^3 \\ \Rightarrow f(-3) = -378 \\ } {/eq}

{eq}\displaystyle{f(0) = 6(0)^5 - 40(0)^3 \\ \Rightarrow f(0) = 0 \\ } {/eq}

To find the local maxima and minima of f(t)

{eq}\displaystyle{f'(t) = \frac{d}{dt}(6t^5 - 40t^3) \\ \Rightarrow f'(t) = 30t^4 - 120t^2 \\ \quad f'(t) = 0 \\ \Rightarrow 30t^4 - 120t^2 = 0 \\ \Rightarrow 30t^2 (t^2 - 4) = 0 \\ \Rightarrow t^2(t^2 - 4) = 0 \\ \Rightarrow t = 0 \ , \ 0 \ , \ 2 \ , \ -2 \\ } {/eq}

Out of these roots, only 0 and -2 lie in the given interval

{eq}\displaystyle \therefore t = 0 \ , \ -2 \\ {/eq}

Second derivative test :

{eq}\displaystyle{f''(t) = \frac{d}{dt} (f'(t)) \\ \Rightarrow f''(t) = \frac{d}{dt} (30t^4 - 120t^2) \\ \Rightarrow f''(t) = 120 t^3 - 240 t \\ } {/eq}

{eq}\displaystyle{ \therefore f''(0) = 0 \\ and, \ f''(-2) = 120(-2)^3 - 240(-2) \\ \Rightarrow f''(-2) = -480 <0 \\ } {/eq}

Therefore, t = -2 is a point of local maximum.

{eq}\displaystyle{f(-2) = 6(-2)^5 - 40(-2)^3 \\ \Rightarrow f(-2) = 128 \\ } {/eq}

Comparing this value with f(0) and f(-3), we conclude that :

{eq}\displaystyle{ \text{Absolute maximum value} = f(-2) = 128 \\ \text{Absolute minimum value} = f(-3) = -378 \\ } {/eq}

#### Learn more about this topic:

Finding Minima & Maxima: Problems & Explanation

from General Studies Math: Help & Review

Chapter 5 / Lesson 2
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