Question

calculus

Answer 1

\[\large\rm g(x)=\int\limits_{-1}^x f(t)dt\]We would like to find a `maximum` for g(x).
This will correspond to some critical point of g(x), yes?
g'(x)=0 <-- to find maximum.

So first, by using the FTC, Part 1, do you understand how to find g'(x)? :)

Answer 2

yes, I think so

Answer 3

you just plug in x into f(t) and the derivative of x is just 1

Answer 4

Yes, you plug x into f(t), good.
And nothing extra from chain rule since it was just an x.
\[\large\rm g'(x)=f(x)\]From here we can use our chart to figure out what is going on.

Answer 5

that's the part i'm stuck on. do i only look at f'(x) since i'm finding a relative maximum?

Answer 6

It's a little strange :)
But we're saying that the derivative of g(x) is the same as f(x).
Not that g(x) is the same as f(x).

So we'll have to be careful in our chart.

Answer 7

So we're looking for critical points,
\(\large\rm g'(x)=f(x)=0\)

So what do we have for `candidates`?
What possible values could be our maximum?
Look for f(x)=0

Answer 8

-1 and 2

Answer 9

Mmm k good good good.

Answer 10

Now think about g'(x).
It has to be `positive` then `negative` to create a hilltop, yes?
That would correspond to a maximum.

So again, don't be tempted into looking at f'(x).
We don't want anything to do with him :)

g'(x) is our f(x).

Answer 11

So we want to know where f(x) goes from `positive` to `negative`

Answer 12

Any ideas? :)
Still confused?

Answer 13

x=2

Answer 14

Ah, yes.
Good job! :)

Answer 15

Thank you! Do you mind helping me with one more question?

Answer 16

sure!

Answer 17

\[\large\rm f(x)=x^2,\qquad g(x)=\sin x,\qquad h(x)=\int\limits_{-1}^{g(x)}f(t)dt\]Hmm these problems are neat :D
Very sneaky questions.

Answer 18

yes, yes they are
I'm confused whether sinx goes into f(t) or x^2 does

Answer 19

Before dealing with the pi/6 business,
let's find our h'(x) using FTC, Part 1.

Answer 20

Any ideas what that will look like? :)
It's going to be a little messy since our upper bound is more than just an x.

Answer 21

Sorry I sort of ignored what you said because we really would like to avoid plugging anything in at this point :))
It will make things way easier for us, hopefully.

Answer 22

sorta
\[\int\limits_{-1}^{sinx}f(t)dt\]
h'(x)=sin(x)^2
This is what i think it'll look like

Answer 23

No no no XD See you plugged stuff in, bahhh :D lol

Answer 24

Ok you were close though :))

Answer 25

\[\large\rm h(x)=\int\limits_{-1}^{g(x)}f(t)dt\]FTC, Part 1 gives us,\[\large\rm h'(x)=f(g(x))\cdot g'(x)\]Take a look at that a sec :O
Too confusing?

The f(g(x)) is actually the part you came up with.
We just forgot that we have to chain rule when the upper bound is more than an x.

Answer 26

so i just have to take the derivative of sinx^2?

Answer 27

g(x)=sin x
So it looks like it's simpler than that.
g'(x)=cos x

Answer 28

aren't we taking the derivative of sin^2x instead of sinx?

Answer 29

\[\large\rm h(x)=\int\limits_{-1}^{stuff} f(t)dt\]Integrating leads to some anti-derivative function, let's call it F(t).\[\large\rm h(x)=F(t)|_{-1}^{stuff}\]Let's plug in our limits,\[\large\rm h(x)=F(stuff)-F(-1)\]Taking derivative brings us back to little f,
but now we have some chain rule going on,\[\large\rm h'(x)=f(stuff)stuff'-0\]So no,
this process doesn't require us to take the derivative of the "entire thing",
only the stuff that was the upper bound.

Answer 30

ooh, so h'(x)=-sinxcosx

Answer 31

\[\large\rm h'(x)=f(g(x))\cdot g'(x)\]\[\large\rm h'(x)=f(\sin x)\cdot (\sin x)'\]\[\large\rm h(x)=(\sin x)^2\cdot (\cos x)\]

Answer 32

oh ok, I see. So now you plug in pi/6 into the x's

Answer 33

Yes :)

Answer 34

That last line should be h'(x), I made a typo.
Hopefully you noticed that though :D

Answer 35

\[\sqrt{3}/8\]

Answer 36

yay good job!

Answer 37

YAY!! Thank you so much for the help!! :D

Answer 38

np c: