Douglas Kraatz

2021-11-14

Let $f\left(x\right)={x}^{3}-6{x}^{2}-15x+23$
a) Perform a full sign analysis on $f\left(x\right):$ find its zeroes, and determine its sign on each interval.
b) Use the information in (2) to determine intervals of concave up / concave down for f.
c) Find the coordinates (x, y) of the inflection point(s) of f.
d) Find the absolute maximum and minimum values of f on the interval . Justify your answer with appropriate calculus work.

Ruth Phillips

Step 1
Given:
$f\left(x\right)={x}^{3}-6{x}^{2}-15x+23$
${f}^{\prime }\left(x\right)=3{x}^{2}-12x-15$
a) $f{}^{″}\left(x\right)=6x-12=0⇒x=2$
b) Concave up on interval
Concave dow on interval
c) Cooedinate of inplection point
$F\left(2\right)={2}^{3}-6×{2}^{2}-15×2+23$
$=8-24-30+23$
$=-23$

d) ${F}^{\prime }\left(x\right)=0⇒3{x}^{2}-12x-15=0⇒{x}^{2}-4x-5=0$
${x}^{2}+x-5x-5=0$
$x\left(x+1\right)-5\left(x+1\right)=0$
$\left(x+1\right)\left(x-5\right)=0⇒x=-1$ and $x=5$
$F\left(-1\right)={\left(-1\right)}^{3}-6×{\left(-1\right)}^{2}-15×-1+23$
$=-1-6+15+23=31$
$F\left(5\right)={5}^{3}-6×{5}^{2}-15×5+23=125-150-75+23=-77$
$F\left(7\right)={7}^{3}-6×{7}^{2}-15×7+23=343-294-105+23=-33$
$F\left(0\right)={0}^{3}-6×{0}^{2}-15×0+23=0+0+0+23=23$
Absolute maxima at $x=-1$ and maximum volue is 31
Absolute minima at $x=5$ and minimum volue is $-77$

Do you have a similar question?