Question

Find the absolute max value and absolute min value of the function.
(a) \[ g(x)=x^2+\frac{ 16}{ x } , 1\le x \le 4\]
(b)\[f(x)=\left( x^2-9 \right)^{2/3}, -6\le x \le6\]

Answer 1

\[\Large\rm g(x)=x^2+\frac{16}{x},\qquad x\in[1,4]\]We'll write g(x) like this,\[\Large\rm g(x)=x^2+16x^{-1},\qquad x\in[1,4]\]So again, I applied an exponent rule to make it look like that.

Answer 2

This should make it easier to apply our power rule, yes?

Answer 3

Yes

Answer 4

So what do we get for g'(x)?

Answer 5

g'(x)=2x-16x^-2

Answer 6

\[\Large\rm g'(x)=2x-16x^{-2}\]\[\Large\rm g'(x)=2x-\frac{16}{x^2}\]Ok good. Let's get a common denominator,\[\Large\rm g'(x)=\frac{2x^3-16}{x^2}\]

Answer 7

We'll look for critical points,\[\Large\rm 0=\frac{2x^3-16}{x^2}\]

Answer 8

The denominator actually gives us some information, just not what we're looking for.
It tells us where we can find `vertical tangents`, but we want `horizontal tangents` to the curve.

Answer 9

So that leads us to,\[\Large\rm 0=2x^3-16\]

Answer 10

Solve for x, what do you get? :o

Answer 11

x=2!

Answer 12

Ok good good good.

Answer 13

Recall that when we're in a closed interval, we also need to check the endpoints!

Answer 14

From here it's pretty simple:
We'll plug the end points into the function,
we'll plug the critical point into the function,

largest output = our maximum
smallest output = our minimum

Answer 15

\[\Large\rm g(x)=x^2+\frac{16}{x}\]

\[\Large\rm g(1)=?\]\[\Large\rm g(2)=?\]\[\Large\rm g(4)=?\]

Answer 16

g(1)= 17
g(2)=12 (absolute min)
g(4)=20 (absolute max)

Answer 17

mmm ok good job! :)

Answer 18

Need some help on the second one, please.

Answer 19

\[\Large\rm f(x)=\left( x^2-9 \right)^{2/3}, \qquad x\in[-6,6]\]Were you able to find a derivative ok?

Answer 20

\[f'(x)=2/3(x^2-9)^{-1/3}(2x)\]
I'm not sure if this is correct?

Answer 21

\[\Large\rm f'(x)=\frac{2}{3}(x^2-9)^{-1/3}(2x)\]Mmmm ok looks good.

Answer 22

Let's condense it down a tad bit,\[\Large\rm f'(x)=\frac{4x}{3(x^2-9)^{1/3}}\]

Answer 23

Oh a note about the last problem, because it's going to affect what we do here...

Answer 24

I should not have thrown out the denominator so quickly in the last problem.
We should have considered that point.

It would have led to a different type of critical point at x=0.
But since x=0 was NOT in our interval 1 to 4, we could easily throw it away.
But not for the reason I mentioned before.

Answer 25

So in this problem, we'll look for critical points as we did before,\[\Large\rm 0=\frac{4x}{3(x^2-9)^{1/3}}\]Let's just start by multiplying both sides by 3/4,\[\Large\rm 0=\frac{x}{(x^2-9)^{1/3}}\]

Answer 26

We want to look at the denominator and numerator separately.
Setting the denominator equal to 0 will give us these "other" types of critical points,\[\Large\rm 0=(x^2-9)^{1/3}\]
What values of x do we get here?

Answer 27

x=3?

Answer 28

Mmm well you eventually get to,\[\Large\rm 0=x^2-9\]\[\Large\rm 9=x^2\]And to isolate x we need to square root,
when we square root a square, we get the ole fancy plus/minus yes?

Answer 29

The alternative is to apply your difference of squares rule from this step,\[\Large\rm 0=x^2-9\]\[\Large\rm 0=(x-3)(x+3)\]

Answer 30

oh yeah! Forgot about that!

Answer 31

So we get two values for x, x=-3, x=3.
Now we'll set the numerator equal to zero and see what's going on there.

Answer 32

That gives us a nice simple value x=0, yah?

Answer 33

yes

Answer 34

So again, we'll plug our `end points` into the function,
we'll plug our `critical points` into the function,
and compare.

Answer 35

\[\Large\rm f(x)=\left( x^2-9 \right)^{2/3}\]

\[\Large\rm f(-6)=?\]\[\Large\rm f(-3)=?\]\[\Large\rm f(0)=?\]\[\Large\rm f(3)=?\]\[\Large\rm f(6)=?\]

Answer 36

Here is what the function looks like graphed,
https://www.desmos.com/calculator/rfuub7idg0
Spoiler Alert though. hehe

Answer 37

f(-6)=9
f(-3)=0

Answer 38

f(0)=4.3...
f(3)=0

Answer 39

f(6)=9

Answer 40

Okay, thank you so much, zepdrix! :)

Answer 41

I finally finished my math homework! YAY!!

Answer 42

yay \c:/