prsategazd

2021-12-14

Given $f\left(x\right)=2{x}^{3}+15{x}^{2}-36x+9$ Find the absolute maximum and the absolute minimum values of this function.

Robert Pina

Step 1
Given function is :
$f\left(x\right)=2{x}^{3}+15{x}^{2}-36x+9$
To find the absolute maximum and the absolute minimum values of this function.
Differentiate the function with respect to x ,
${f}^{\prime }\left(x\right)=6{x}^{2}+30x-36$
Step 2
For maxima or minima f'(x)=0 , therefore
$6{x}^{2}+30x-36=0$
$6\left({x}^{2}+5x-6\right)=0$
(x+6)(x-1)=0
x=1, -6
Now differentiate f'(x) with respect to x,
f''(x)=12x+30
f''(x) at x=1
$f{}^{″}\left(1\right)=12×1+30$
=42>0
f''(1)>0, therefore f(x) has minimum value at x=1
Step 3
At x=1 , f(x) has minimum value and corresponding minimum value of f(x) is
$f\left(1\right)=2×1+15×1-36×1+9$
=-10
Now
f''(x) at x=-6
$f{}^{″}\left(-6\right)=12×\left(-6\right)+30$
=-42<0
f''(-6)<0, therefore f(x) has maximum value at x=-6
Step 4
At x=-6 , f(x) has maximum value and corresponding maximum value of f(x) is
$f\left(-6\right)=2×{\left(-6\right)}^{3}+15×{\left(-6\right)}^{2}-36×\left(-6\right)+9$
=333
Thus, the
Maximum value of f(x) is 333
Minimum value of f(x) is (-10)

Do you have a similar question?