# Thread: Maxima, minima points of inflection

1. ## Maxima, minima points of inflection

Hi not sure if I've got this right.

Determine any points at which the gradient of $y=x^3-3x^2-1$ is zero, and detemine whether these are maxima, minima or points of inflection.

$3x^2 - 6x$
$3x(x-2)$

x = 2

$(2)^3-3(2)^2-1$

(2, -5)

(left of turning point) x = 1

$3(1)^2-6(1)= -3$

(1, -3)

(right of turning point) x = 3

$3(3)^2-6(3)= -9$

(1, -9)

I'm not sure if I've done this right, but if I were to answer this I believe (2, -5) is minima.

2. ## Re: Maxima, minima points of inflection

What you've done is almost fine but you have to be more clear and careful with mathematical expressions like:
Originally Posted by Dcoz
$3x^2 - 6x$
$3x(x-2)$

x = 2
When you read it like this, this doesn't make sense although you mean:
$3x^2-6x=0 \Leftrightarrow 3x(x-2)=0 \Leftrightarrow 3x=0 \ \mbox{or} \ x-2=0$
So we have two zero's $x=0$ and $x=2$

To determine if you're dealing with a minimum or a maximum you can make a sign table (I guess you've done something like that):
You get a maximum $(0,-1)$ and a minimum $(2,-5)$ (review this)

To determine if there're points of inflection then you've to use the second derivative ...

3. ## Re: Maxima, minima points of inflection

Thank you. I got (0, -1) for the maximum, but wasn't certain. Thank you also for correcting me on how I displayed the working out.

4. ## Re: Maxima, minima points of inflection

You're welcome! Have you found the points of inflection?

5. ## Re: Maxima, minima points of inflection

I honestly thought that was the question 100% answered due to there being no points of inflection?

6. ## Re: Maxima, minima points of inflection

There's one point of inflection:
$\frac{d^2y}{dx^2}=\frac{d(3x^2-6x)}{dx}=6x-6$
To determine if there're inflection points let $\frac{d^2y}{dx^2}=0$ so:
$6x-6=0 \Leftrightarrow x=1$

If we enter $x=1$ in the cubic function we get for the $y$-coordinate:
$y=-3$

So the point of inflection is:
$(1,-3)$

7. ## Re: Maxima, minima points of inflection

Thank You! You've been a great help.

8. ## Re: Maxima, minima points of inflection

You're welcome!
Also take a look at the graph of the function to compare with the answers you found (you can look at the graph @wolphram alpha or just enter the function in your calculator).

9. ## Re: Maxima, minima points of inflection

I only have a scientific calculator. I'm currently studying engineering product design and didn't believe I would require a graphing calculator.

10. ## Re: Maxima, minima points of inflection

Ok, then you can take a look @ wolphram alpha (if you want), it's a very useful site.

11. ## Re: Maxima, minima points of inflection

I've started using it quite a bit now. Just wish it could help with my physics questions :-(

12. ## Re: Maxima, minima points of inflection

There's also the PHF for physics questions .

13. ## Re: Maxima, minima points of inflection

Yeah, but unfortuately they don't seem to be as active as the friendly people over here and for that I thank you again!

14. ## Re: Maxima, minima points of inflection

You're welcome!

15. ## Re: Maxima, minima points of inflection

Be careful, just checking a point on either side of a critical point and seeing that they are lower than the value at the critical point does NOT prove that the critical point is a maximum! For example, if $f(x)= -x^4+ \frac{15}{4}x^2+ 1$, $f'= -4x^3+ \frac{15}{2}x= -4x(x^2- \frac{5}{2})$ so x= 0 is a critcal point. Now f(2)= f(-2)= -16+ 15= -1 while f(0)= 1. But (0, 1) is in fact, a local minimum, not maximum.

We should use, instead, either the "first derivative test", checking that if x< 0, but close to 0, $f'(x)= -4x(x^2- \frac{5}{2})$ we have three negative factors so that f'(x) is negative, while if x> 0 but close to 0, $f'(x)= -4x(x^2- \frac{3}{2}$ has two negative factors (-4and $x^2- \frac{3}{2}$) but one positive factor (x) and so f'(x) is positive. f is decreasing to left of 0 and then increasing- x= 0 is a minimum, or the "second derivative test", seeing that the second derivative is $f''(x)=-12x^2+ \frac{15}{4}$ so that $f''(0)= \frac{15}{4}> 0$. That tells us that the first derivative is increasing at x= 0. Since the first derivative was 0 there, it must be negative to the left an positive to the right- i.e. (0, 1) is a minimum.

For your example, $f(x)= x^3- 3x^2- 1$, $f'(x)= 3x^2- 6x= 3x(x- 2)$ so that, as you say, x= 0 and x= 2 are critical points. We can see that if x< 2, but close to 2 (so x> 0) 3 and x are positive but x- 2 is negative. That means that f'(x) is negative to the left of x= 2 so the function is decreasing toward x. if x> 2 then x> 0 also so all three factors are positive. f' is positive to the right of x= 2 which means that the function is increasing away from x= 2. That tells us that x= 2 gives a minimum. Or we could use the "second derivative test": f''(x)= 6x- 6 so f''(2)= 12- 6= 6> 0. Again, that tells us that x- 2 gives a minimum.

Of course, x= 0 is also a critical point. If x< 0 then x is also less than 2 so we have one positive factor (3) and two negative factors (x and (x- 2)) so f'(x) is positive for x< 0. For x> 0 but close to 0 (so x< 2) 3 and x are positive but x- 2 is still negative. That means that f'(x) is negative. We have f increasing toward x= 0 and decreasing awy from it- x= 0 gives a maximum. Or, f''(0)= 6(0)- 6= -6< 0 so, again, x= 0 gives a maximum.