# Thread: #5) Find the x-value of all pints where the function has relative extrema...

1. ## #5) Find the x-value of all pints where the function has relative extrema...

Find the x-value of all points where the function has relative extrema. Find the values of any relative extrema.

f(x)=x^3-12x+2
y'=3x^2-12=0
y'=3x^2=12
y'=x^2=4

Now after this point what i did was take the square root of each side. Now aren't i supposed to end up with +/- 2 and find the matching y-values? I get:
Max:0@-2
Min:0@2
But the computer say this is wrong i know that 0@2 is right but where did i go wrong?

2. Originally Posted by plstevens
Find the x-value of all points where the function has relative extrema. Find the values of any relative extrema.

f(x)=x^3-12x+2
y'=3x^2-12=0
y'=3x^2=12
y'=x^2=4

Now after this point what i did was take the square root of each side. Now aren't i supposed to end up with +/- 2 and find the matching y-values? I get:
Max:0@-2
Min:0@2
Mr F says: I have no idea what the above notation is meant to mean.

But the computer say this is wrong i know that 0@2 is right but where did i go wrong?
Stationary points at x = 2 or x = -2.

Minimum turning point at x = 2 and maximum turning point at x = -2.

3. min and max are for the relative maximum, and the relative minimum

4. Originally Posted by plstevens
min and max are for the relative maximum, and the relative minimum
I am aware of that. I was referring to the 0@-2 and 0@2 business. Try to use standard mathematical notation if you want people to make sense of what you post.

5. well that's how my teacher posted it, i was very unaware that this was incorrect, sorry if i confused you, but are you able to help

6. it just means 0 at 2, 0 being y and 2 being x

7. Originally Posted by plstevens
well that's how my teacher posted it, i was very unaware that this was incorrect, sorry if i confused you, but are you able to help
$\displaystyle f(2) = 2^3 - 12(2) + 2 \neq 0$. The minimum value at x = 2 is NOT 0.