Question

sketch the following curve, indicating all relative extreme points and inflection points. y = \frac{1}{3}x^{3}+3x^{2}+5x + 1
find the first derivative of y. y=1x^{2}+6x + 5
the relative extreme points are (-5,\frac{28}{3}),(-1,-\frac{4}{3}) (type an ordered pair. simplify your answer. use integers or fractions for any numbers in the expre comma to separate answers as needed.)
the inflection points are (type an ordered pair. simplify your answer. use integers or fractions for any numbers in the express comma to separate answers as needed.)

Explanation:

Step1: Find the second - derivative

Given $y'=x^{2}+6x + 5$, then $y'' = 2x+6$.

Step2: Set the second - derivative equal to zero

Set $y''=0$, so $2x + 6=0$.
Solve for $x$:
$2x=-6$, then $x=-3$.

Step3: Find the $y$ - value of the inflection point

Substitute $x = - 3$ into the original function $y=\frac{1}{3}x^{3}+3x^{2}+5x + 1$.
$y=\frac{1}{3}(-3)^{3}+3(-3)^{2}+5(-3)+1$
$y=\frac{1}{3}(-27)+3\times9-15 + 1$
$y=-9 + 27-15 + 1$
$y=4$.

Answer:

$(-3,4)$