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 (3,13),(5,\frac{35}{3}) (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.) the inflection 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 the second - derivative

Given $y'=x^{2}-8x + 15$, differentiate it with respect to $x$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y'' = 2x-8$.

Step2: Set the second - derivative equal to zero

To find the inflection points, set $y''=0$. So, $2x - 8=0$.

Step3: Solve for $x$

Add 8 to both sides of the equation $2x - 8=0$: $2x=8$. Then divide both sides by 2, we get $x = 4$.

Step4: Find the $y$ - value

Substitute $x = 4$ into the original function $y=\frac{1}{3}x^{3}-4x^{2}+15x - 5$.
\[

$$\begin{align*} y&=\frac{1}{3}(4)^{3}-4(4)^{2}+15(4)-5\\ &=\frac{64}{3}-64 + 60-5\\ &=\frac{64}{3}-9\\ &=\frac{64 - 27}{3}\\ &=\frac{37}{3} \end{align*}$$

\]

Answer:

$(4,\frac{37}{3})$