What is the method for finding inflection points for a given function?

[DMOC]

Homework Statement

Let f be the function f(x) = (x² - 3)e^x for all real #'s x.

a. What values of x is f increasing?
b. What are the x-coordinates of the inflection points for f?
c. Where (by finding the x and y coordinates of a point) is f(x) at its absolute maximum.

Homework Equations

Product rule: uv' + vu'

The Attempt at a Solution

PART A

f(x) = (x² - 3)e^x
f '(x) = e^x(x² + 2x -3)

x = 1 and x = -3

If x < -3, f '(x) is positive which means that f is increasing.
Same thing for x > 1

:. f increases when x < -3 and x > 1

PART B

This is the part where I encounter trouble. Usually I have no problem with these types of problems ... but I can't find when the second derivative equals zero.

f '(x) = e^x(x² + 2x -3)
f ''(x) = e^x(x² +4x -1)

This is a no-calculator-at-all problem so it has to be relatively easy to find when f ''(x) = 0. Here, it's not. Did I make a mistake somewhere?

 

[miqbal]

Use the quadratic formula

x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

No mistakes so far...

 

[DMOC]

Use the quadratic formula

x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

No mistakes so far...

Probably should have done that. I have a habit of viewing the quadratic formula as finding values with lots of decimals but don't round to a fraction. Definitely not the case.

Anyway, here it is.

x = -2 + (root20)/2
x = -2 - (root20)/2

Moving on to part c.

*The absolute maximum point must either be at x=-3 or beyond the point where x=1.

If x = -3, then ...

f(-3) = 6e^-3

If x = a number above 1 (maybe 5)

f(5) = 22e⁵

f(5)>f(-3)

Having greater values of x will only yield higher results; therefore there is no absolute maximum.

Do you think this is ok as a final answer or should I add on to this?

 

[DMOC]

Anyone have their opinions on what I should add to the answer?

 

[LCKurtz]

Remember that a point where f''(x) = 0 is only a potential inflection point. To actually be one, you must show a change in concavity there. You can usually do this by observing that f''(x) changes sign at that point.

Not every point where f''(x) = 0 is an inflection point. Consider f(x) = x⁴ at x = 0.