# Finding a point of inflection

1. Oct 28, 2009

### DMOC

1. The problem statement, all variables and given/known data

Let f be the function f(x) = (x2 - 3)ex 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.

2. Relevant equations

Product rule: uv' + vu'

3. The attempt at a solution

PART A

f(x) = (x2 - 3)ex
f '(x) = ex(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) = ex(x2 + 2x -3)
f ''(x) = ex(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?

2. Oct 28, 2009

### miqbal

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

No mistakes so far...

3. Oct 28, 2009

### DMOC

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?

4. Oct 30, 2009

### DMOC

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

5. Oct 30, 2009

### 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.