Points of Inflection help

The first differential is used to give the point of zero slope. The second derivative determine=s if that point of zero slope is a maxima (positive value), minima (negative value), or inflection (zero)

I'll work through A

f(x)=x^3-x^2

df/dx=3x^2-2x

Set slope (df/dx) equal to zero

3x^2-2x=0
x(3x-2)=0
x=0 & x=2/3

So f(x) has zero slope at x=0 & x=2/3

2nd derivative of f(x) will tell if this is an inflection point
d^2f/dx^2 (f")=6x-2
At x=0 d^2f/dx^2=-2 this means at x=0 f(x) is a maxima
At x=2/3 d^2f/dx^2=0 this is an inflection point

To find the value of y=f(x), just plug the value of x (2/3) at the inflection point into the original function.

f(2/3)=(2/3)^2- (2/3)^2=8/27-4/8=8/27-12/27=-4/27

So the inflection point is (2/3,-4/27)

The rest of the problems can be solved by a like application of the first and second derivatives and a little algebra.

Rap

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Thanks for the walkthrough.

I worked your example and got the same answer, however, my choices are only 1/3, -1/3, -2/27, 2/27, -2/9, 2/9, 2/3 and -2/3.

Can -4/27 be simplified? I can't figure how.

Also, with B, the f'(x) is 7x^6-2x. How do I factor this? Is it x(6x^6-2)?

Then solve for x...?

4 and 27 are relatively prime [-4/27=-(2*2)/(3*3*3)] and have no factors in common. so the answer is no

f(x)=x^7-x^2
f'(x)=7x^6-2x=x(7x^5-2)=0
so one point is obvious (x=0),
the rest are under 7x^5-2=0 or x^5=2/7 & x=(2/7)^(1/5) these are the points where the slope of f(x) is zero.

Rap