Question

1. for what values of $x$ does the graph of $y = 3x^{5} + 10x^{4}$ have a point of inflection?

a $x = \frac{-8}{3}$ only
b $x = -2$ only
c $x = 0$ only
d $x = 0$ and $x = \frac{-8}{3}$
e $x = 0$ and $x = -2$

Explanation:

Step1: Find first derivative

$y' = \frac{d}{dx}(3x^5 + 10x^4) = 15x^4 + 40x^3$

Step2: Find second derivative

$y'' = \frac{d}{dx}(15x^4 + 40x^3) = 60x^3 + 120x^2$

Step3: Solve $y''=0$ for $x$

$60x^3 + 120x^2 = 0$
$60x^2(x + 2) = 0$
Solutions: $x=0$, $x=-2$

Step4: Test concavity change

• For $x < -2$ (e.g., $x=-3$): $y''=60(-3)^3+120(-3)^2=-540+1080=540>0$ (concave up)
• For $-2 < x < 0$ (e.g., $x=-1$): $y''=60(-1)^3+120(-1)^2=-60+120=60>0$ (concave up, no change at $x=0$)
• For $x > 0$ (e.g., $x=1$): $y''=60(1)^3+120(1)^2=180>0$ (concave up, no change at $x=0$)
• At $x=-2$: Concavity changes from up to up? Correction: Re-test $x=-3$: $y''=60(-27)+120(9)= -1620+1080=-540<0$ (concave down). $x=-1$: $y''=-60+120=60>0$ (concave up). So concavity changes at $x=-2$, not at $x=0$.

Answer:

B. $x = -2$ only