Inflection Points

[Becky4paws]

Determine where the given function is increasing and decreasing and where its graph is concave up and concave down. Find the relative extrema and inflection points.

f(x) = 2s(s+4)^3

f'(x) = 6s(s+4)^2
critical points: (-4,0), (0,0)

f"(x) = 12s(s+4)
critical points/inflection points: (-4,0), (0,0)

for x<-4: f'(x) value is positive, f"(x) value is positive, increasing and concave up.

-4<x<0: f'(x) value is negative, f"(x) value is negative, decreasing and concave down.

x>0: f'(x) value is positive, f"(x) value is positive, increasing and concave up.

I charted the graph but don't see how why point (-2, -32) is an inflection point.

 

[galactus]

Determine where the given function is increasing and decreasing and where its graph is concave up and concave down. Find the relative extrema and inflection points.

Your derivatives are incorrect.

f(x) = 2s(s+4)^3

f'(x) = 6s(s+4)^2
critical points: (-4,0), (0,0)

\(\displaystyle f'(x)=8(x+1)(x+4)^{2}\)

f"(x) = 12s(s+4)
critical points/inflection points: (-4,0), (0,0)

\(\displaystyle f''(x)=24(x+2)(x+4)\)

for x<-4: f'(x) value is positive, f"(x) value is positive, increasing and concave up.

-4<x<0: f'(x) value is negative, f"(x) value is negative, decreasing and concave down.

x>0: f'(x) value is positive, f"(x) value is positive, increasing and concave up.

I charted the graph but don't see how why point (-2, -32) is an inflection point.

See the change in concavity at x=-2?. Therefore, a inflection point.