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

• Sep 24th 2008, 12:49 PM
plstevens
#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?
• Sep 24th 2008, 01:27 PM
mr fantastic
Quote:

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.
• Sep 24th 2008, 01:32 PM
plstevens
min and max are for the relative maximum, and the relative minimum
• Sep 24th 2008, 01:39 PM
mr fantastic
Quote:

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.
• Sep 24th 2008, 01:55 PM
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
• Sep 24th 2008, 01:56 PM
plstevens
it just means 0 at 2, 0 being y and 2 being x
• Sep 24th 2008, 02:01 PM
mr fantastic
Quote:

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

• Sep 24th 2008, 02:06 PM
plstevens
i think its relative minimum of 0 at 2, thats how it is in the anwer choice
• Sep 24th 2008, 02:21 PM
mr fantastic
Quote:

Originally Posted by plstevens
i think its relative minimum of 0 at 2, thats how it is in the anwer choice

$f(2) = 2^3 - 12(2) + 2 \neq 0$. The minimum value at x = 2 is NOT 0.
• Mar 19th 2011, 11:14 AM
HallsofIvy
The problem, as you posted it, was "Find the x-value of all points where the function has relative extrema". Those x values are -2 and 2. The problem did not ask for the corresponding y values.