Question

How would I find the relative extrema of f(x)=1/4(x+2)(x-1)^2?

Answer 1

You get intercepts by placing either x=0 (where your graph hits the Y axis) or y=0 (hits on X axis), and solving for the remaining variable.
relative extrema are the point where the 1st order derivative is zero and changes sign around it).

Answer 2

well i got the intercepts, but im still not sure how to find the relative extrema :(

Answer 3

i dont understand it at all, like how would i find the 1st order derivative?

Answer 4

Errr, you aren't studying calculus, are you?

Answer 5

pre-calculus, but math is definitely not my strongest :o

Answer 6

Problem gets trivial once you know what derivatives are, it gets mechanical. What's the last topic you've seen?

Answer 7

polynomials and rational functions :o

Answer 8

Call me spoiled (or tired), but I don't see a way to find extrema without derivatives.

Answer 9

I don't blame you! Haha my head hurts from all of this math already! But would the derivative be d/dx(1/4 (-1+x)^2 (2+x)) = 3/4 (x^2-1)? I used wolframalpha

Answer 10

But even if that is the derivative, I don't know how to find the relative extrema

Answer 11

Trusting its calculation (lazy mode ON) you need to see how its sign varies when x varies. Easiest way to do it is with a graph: you get where d/dx = 0, and if around this/theese point it changes sign (ie <0, 0, >0 or vice versa) you have a relative minimum or maximum.

Answer 12

hm, well these are the graphs it gave also

Answer 13

does that help narrow it down?

Answer 14

That's being spoiled, but look at your derivative: you know how to analize a sign of a function?

Answer 15

i'm not sure, i've never done that before

Answer 16

First, you draw a line, which rapresents your real axis (x). You mark in it every number where your function* either doesn't exist (which isn't your situation) or is zero.
*in this case, the derivative 3/4 (x^2-1)

In the specific case, it's zero for x=1 or x= -1.

Answer 17

As you see, you are splitting your axis in as many parts as your zeroes, plus one. For each zone your function has the same sign whatever number you pick: still in this case, on both -1.000001 (just a bit less than -1) and -10000001 (WAAAY less than -1) you get the same sign. Trick is: you pick convenient values of x in each interval and see how your sign goes.
ie, for x< -1 you pick -2 (the cloesest integer),
for -1 < x < 1 you pick 0
for x > 1 you pick 2. You could do the calculation with -230, 43/52 and 999 for each range, but i doubt you like to suffer more than needed. Your graph becomes something like
And once you see that for both -1 and 1 your derivative IS 0 and CHANGES SIGN around each point each are extrema.

If there is a method which allows you not to use derivatives... be my guest to find it.

Answer 18

Oh I see, so the relative extrema would be 1 and -1?

Answer 19

Yes, more properly (-1, 1) and (1;0)

Answer 20

thank you so much!:) do you think you could help me find the relative extrema of h(x)=2x^3+5x^2-25x?

Answer 21

\[h'(x) = 2*3x^2 +5*2x -25 = 6x^2 -10x -25\]
\[h'(x) = 0 \rightarrow x= {{5 \pm 5\sqrt7}\over 6}\]
outside theese values h'(x) > 0, inside it's negative (graph is exactly the same as above, only number changes).

Always do the "do the sign change" test because you can have situation like y=x^3 where your derivative is 0 for x =0 but you don't get an extrema (plot it if you're curious)