Question

Identify the extrema for the function. Classify each as a local or absolute max or min value:
f(x)=4X^3+15X^2-8X-3

How does one do this?

Answer 1

First step is to differentiate.

Answer 2

What do you mean?

Answer 3

Hmmm... so this isn't a calculus question?

Answer 4

It's pre-calc haha. I know what the graph should look like, but don't know how to algebraically get to the max and min values.

Answer 5

Well the only way I know how to do it without calculator is to differentiate. But you don't know how to differentiate so I don't know.

Answer 6

I just looked it up just to make sure I wasn't just having a brain freeze, but yeah we haven't learned that method :/

Answer 7

Thank you though

Answer 8

So the only other way I know is to graph it on your calculator and trace the graph to find the max(s) and min(s).

Answer 9

are you sure you never differentiated before?
like finding f' given f(x)=2x^3 is f'(x)=2*3x^2 by constant multiple and power rule

Answer 10

You can also differentiate f by using the definition of derivative

Answer 11

Nope haha, not yet at least! Yeah when I plug it into my calculator, it looks like the min value is at -4 when X=1/3, but it's hard to tell, so I need an exact answer. I've done the derivative before

Answer 12

wait what?

Answer 13

you said nope and then the last sentence you say yes

Answer 14

that is contradicting information :p

Answer 15

Sorry I meant "I've not found..."

Answer 16

"I've not done the derivative before"

Answer 17

I'm looking up how to though

Answer 18

there is a long way and some short cut ways which the long way proves

Answer 19

teacher usually explains the long way first since it is the definition and also where the short cuts come from but then the class usually starts to use just the short cut ways but remembers the definition of derivative just in case it should come up somewhere

Answer 20

this is how you find the derivative of f:
\[\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)\]

Answer 21

that is the definition of derivative

Answer 22

\[\small{f'(x)=\lim_{h \rightarrow 0} \frac{[4(x+h)^3+15(x+h)^2-8(x+h)-3]-[4x^3+15x^2-8x-3]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{4[(x+h)^3-x^3]+15[(x+h)^2-x^2]-8[(x+h)-x]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{4[(x+h-x)((x+h)^2+(x+h)(x)+x^2)+15(x+h-x)(x+h+x)-8h}{h}} \\ f'(x)=\lim_{h \rightarrow 0} \frac{4h[(x+h)^2+x(x+h)+x^2]+15h(2x+h)-8h}{h} \\ \text{ then recall } \frac{h}{h}=1 \\ \text{ so we have } \\ f'(x)= \lim_{h \rightarrow 0} \frac{4[(x+h)^2+x(x+h)+x^2]+15(2x+h)-8}{1} \frac{h}{h} \\ f'(x)= \lim_{ h \rightarrow 0} (4[(x+h)^2+x(x+h)+x^2]+15(
2x+h)-8 )\]
now since the bottom isn't going to be 0 we can finally plug in 0 since this function is continuous at h=0
\[f'(x)=4[(x+0)^2+x(x+0)+x^2]+15(2x+0)-8 \\ f'(x)=4[3x^2]+15(2x)-8 \\ f'(x)=12x^2+30x-8\]
but like I said there is a short cut way to do this
then you find the critical numbers
if there are any max(s) or min(s) they will occur at the critical points
you can find the critical points for this one just by solving f'(x)=0
or that is solving 12x^2+30x-8=0

Answer 23

Oh my goodness, thank you for typing all of that out for me! I am going to write it down to refer to it when we do start learning it!

Answer 24

now the x values are where the max(s) and min(s) occur
but f will give you the max(s) and min(s) once you find the critical numbers and plug those into f

Answer 25

if you are taking precal I'm almost certain you guys will cover the definition of derivative
i'm surprised you haven't given this question

Answer 26

the way I went about it wasn't the only way to find the limit (or the derivative)
I could have expanded things and cancel but instead I decided to group the cubes together
and the squares together and linears together
and factor the difference of cubes and factor the difference of squares

Answer 27

basically I used these two formulas:
\[a^3-b^3=(a-b)(a^2+ab+b^2) \\ \text{ and } a^2-b^2=(a-b)(a+b)\]

Answer 28

Oh okay, haha it seems so intense! Thank you, for taking the time. I am writing it down, and I am going to ask my teacher about it :)

Answer 29

It will be intense and even may be confusing but with enough practice I believe you can get it and not think it so intimidating.

Answer 30

Yeah, I will definitely look over it and I'm going to check the lesson plan and see if we are learning it soon (I'm virtual schooled, so I can see what we are doing for the next few months)

Answer 31

by the way the local max should be closer to the point (-2.7,49.3)
and the local min should be closer to the point (.2,-4)
these aren't exact values but you can find the exact values by using that quadratic equation I gave you and inputting those values into f to find the corresponding y value

Answer 32

this is a cubic so there is no absolute extrema

Answer 33

only local

Answer 34

Okay, thank you :) I wonder if my teacher just expects us to guesstimate since we never learned that equation. And okay thank you! So you can tell from the degree if there are absolutes or locals?

Answer 35

since we were given no restrictions on the domain
we normally assume all real numbers
and yes you can look at the degree to help you determine the end behavior of the graph
and this is what will help you know if there is a possibly absolute extrema

Answer 36

Alrighty! I know the behaviors of graphs, so that's a plus haha! Thank you so much for everything

Answer 37

np

Answer 38

I don't know how much I helped considering I didn't do it the way you guys were suppose to whatever way that was suppose to be...
I kinda think since you guys haven't actually learned the derivative, it is just a guessimation thing by observing and tracing the graph with your ti83 (or whatever equivalent you have to that)

Answer 39

Yeah, I am going to guesstimate for now and then when I turn it in, I will ask her and mention the derivative :)