How many points of inflection does the function F(x)=(pi/3)^(x^3 - 8) have?

Question

anonymous · Answer

you will have to differentiate twice and set it equal to 0

anonymous · Answer

if you just need the number of inflection points, you can safely ignore the 
$$\frac{\pi}{3}$$ out front. 
$$y=x^3-8$$ looks just like 
$$y=x^3$$translated down 8 units. since 
$$y=x^3$$ has one inflection point at (0,0) 
$$y=x^3-8$$ also has one inflection point, this one at (0,-8)

asnaseer · Answer

unless your function is:$$F(x)=(\frac{\pi}{3})^{x^3-8}$$

asnaseer · Answer

is it the one I displayed above or is it:$$F(x)=\frac{\pi}{3}(x^3-8)$$

anonymous · Answer

No the function is F(x)=$$(\pi/3)^{x^{3}-8}$$

asnaseer · Answer

then I you cannot follow what @satellite73 showed as I think he assumed the other form of F(x).

As @imranmeah91 said, what you need to do is:

find the values for x that make F''(x) = 0

asnaseer · Answer

you can use wolfram to help you: http://www2.wolframalpha.com/input/?i=inflection+points+%28pi%2F3%29%5E%28x%5E3-8%29

anonymous · Answer

Thank You =)

anonymous · Answer

Nice job asnaseer

anonymous · Answer

whoa did i ever read that wrong. sorry!

anonymous · Answer

all wrong again

asnaseer · Answer

no worries @satellite73 - after all - to err is to be human...