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# Find the absolute maximum and absolute minimum values of $f$ on the given interval: $\;f(x)=x^{3}-12x+1 ,[-3 , 5]$

Note: This is part 3rd of a 7 part question, split as 7 separate questions here.

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Toolbox:
• To find the absolute maximum and minimum values of a continuous function f on a closed interval $[a,b]$
• (i) Find the values of the critical numbers of f in $(a,b)$
• (ii) Find the value of $f(a)$ and $f(b)$
• (iii) The largest of the values from (i) and (ii) is the absolute maximun value, the smallest of these values is the absolute minimum value.
$f(x)=x^3-12x+1$ is continuous on $[-4,1]$
Step 1:
$f'(x)=3x^2-12$
$f'(x)=0 =>x^2=4$ or $x =\pm 4$
Step 2:
The critical values are
$f(x)=64-48+1=17$
$f(-4)=-64+48+1=-15$
The values of f at the endpoints are
$f(-3)=-27+36+1=10$
$f(5)=125-60+1=66$
Step 3:
Comparing the 4 values, the absolute maximum is $f(5)=66$ and the absolute minimum is $f(-4)=-15$
answered Jul 31, 2013 by