Question

calculus help. I vive medals

Answer 1

@jim_thompson5910

Answer 2

@Data_LG2

Answer 3

@zepdrix do you think you can help me?

Answer 4

So if you integrate,\[\large\rm \int\limits_{-3}^0 f'(x)dx\quad=\]Can you handle the next step? What would f'(x) integrate to?

Answer 5

6

Answer 6

wait

Answer 7

I'm not sure if this is the correct approach actually :)
But at least it'll get us thinking lol

Answer 8

nvm, 9/2

Answer 9

9/2, ok great.
You must've done that by counting the squares.

But let's write the anti-derivative of f'(x) as f(x), yes?\[\large\rm \int\limits\limits_{-3}^0 f'(x)dx\quad=f(x)|_{-3}^0\]Evaluating at the limits,\[\large\rm \int\limits_{-3}^0 f'(x)dx\quad=f(0)-f(-3)\]

Answer 10

From there we can plug in some information.

Answer 11

ok

Answer 12

As you determined, we can replace the integral with the value 4.5.\[\large\rm 4.5=f(0)-f(-3)\]And they also told us that f(0)=7,\[\large\rm 4.5=7-f(-3)\]

Answer 13

do we just subtract minus 7 and then multiply by negative?

Answer 14

in that case, I got 2.5

Answer 15

Yes, that seems like a good idea.
That will give us the value of f(-3).

Try to think about the process for determining a minimum on an interval.
You have to check a few things:
~ Find critical points
~ Then Plug any critical points, AND end points, into the function
and compare the outputs.

Smallest output = minimum.

Answer 16

So we have that:
f(0) = 7
f(-3)= 2.5

Any critical points in [-3,0] that we should look at?
Why or why not? :)

Answer 17

no because if we had critical points, then our value would have been 0

Answer 18

it seems that as you increase x, your value increase. It appears reasonable.

Answer 19

Good, yes.
The function is strictly increasing on [-3,0] because the first derivative is positive over that entire interval. So no critical points.

Simply compare the end points.
Looks like f(-3) gave us the smallest possible value, ya? :)

Answer 20

ya

Answer 21

f(-3) = 2.5

Yayy good job Javier \c:/

Answer 22

Thanks. Would you help with another one? I selected and answer but I am not sure if thats correct

Answer 23

sure!

Answer 24

One sec, lemme upload a small picture so I can write on the graph.

Answer 25

You have a couple options here:

You can take the good student route, and do this the calculus way,
looking at slopes and comparing values.

Or you can sort of cheat,
and realize that they graphed an x^4, x^3 and x^2 type of function.
And you know that polynomials decrease in degree when you differentiate them.
So whatever function looks like x^4 is your f(x).

Answer 26

The calculus way is a lot more helpful though, let's see if you can follow along.

Answer 27

a