Question

sketch the following curve, indicating all relative extreme points and inflection points. y = \frac{1}{3}x^{3}-4x^{2}+15x - 5
find the first derivative of y. y=x^{2}-8x + 15
the relative extreme points are
(type an ordered pair. simplify your answer. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)

Explanation:

Step1: Find critical points

Set $y' = 0$, so $x^{2}-8x + 15=0$.
Factor the quadratic equation: $(x - 3)(x - 5)=0$.
Solve for $x$: $x=3$ or $x = 5$.

Step2: Find the $y$-coordinates of critical points

When $x = 3$, $y=\frac{1}{3}(3)^{3}-4(3)^{2}+15(3)-5=9 - 36+45 - 5=13$.
When $x = 5$, $y=\frac{1}{3}(5)^{3}-4(5)^{2}+15(5)-5=\frac{125}{3}-100 + 75-5=\frac{125}{3}-30=\frac{125 - 90}{3}=\frac{35}{3}$.

Answer:

$(3,13),(5,\frac{35}{3})$