Question

Assume
f ' is given by the graph in the figure. Suppose f is continuous and that
f(0) = 0.

(a) Find
f(3)= .


(b) Find f(7)=


(c) Find
7
(integral)f '(x)dx=
0


(d) Sketch a graph of f over the interval
0 ≤ x ≤ 7.

Answer 1

Other than the answer, what don't you get?

Answer 2

i understand the question; its just that the graph is discontinuous at the points where I am supposed to find… f(3) and f(7) so I would not know how to determine the y value

Answer 3

well if want to think about what happens around x=3 on your graph f'
we need integrate both sides of x=3 as a first step

Answer 4

okay

Answer 5

so what do you get integrating both sides

Answer 6

don't forget to add your constant

Answer 7

after that we want to use that f is continuous so the left limit has to equal the right limit as x approaches 3

Answer 8

so if y'=4 then y=?
and if y'=-2 then y=?

Answer 9

do you know?

Answer 10

sorry for the lag! but no my mind isn't working

Answer 11

i know you can integrate a constant

Answer 12

\[\int\limits_{}^{} c dx=cx+K\]

Answer 13

since the derivative of (cx+K) is c

Answer 14

so if y'=4 then y=?

Answer 15

\[y=\int\limits_{}^{}y' dx\]

Answer 16

4x

Answer 17

4x+constant

Answer 18

so let's just skip to ensuring we have continuity

Answer 19

a and b are constant values

right function to x=3 is y=4x+a
left function to x=3 is y=-2x+b

right limit = left limit (as x approaches 3)
4(3)+a = -2(3)+b solve for either a or b

Answer 20

okay lets see!

Answer 21

the part after this we are going to use that f(0)=0

Answer 22

and then we will able to find f(3)

Answer 23

18+a=b

Answer 24

\[y=-2x+b \\ y=-2x+18+a\]

Answer 25

or f(x)=-2x+18+a
to find a use f(0)=0

Answer 26

where did the -2x come from??

Answer 27

we had to integrate both sides of x=3

Answer 28

you did once side
i did the other side

Answer 29

ohhh okay gotcha

Answer 30

then we set them equal to each other as x goes to 3

Answer 31

and solve for either a or b

Answer 32

then you found b=18+a
then I replace b with 18+a

Answer 33

so then we had y=-2x+b
so y=-2x+18+a

Answer 34

now we are also given f(0)=0
and we are going to use that to find our constant value a

Answer 35

so from there do we plug in 0

Answer 36

yes since we know f(0)=0
f(0)=-2(0)+18+a=0
18+a=0

Answer 37

so -18

Answer 38

a=-18

Answer 39

so we have almost found our function at x=3

Answer 40

yay!!

Answer 41

f(x)=-2x+18+(-18)=-2x
or
you could use
f(x)=4x+a=4x-18

either one of these should given you the same value at x=3

Answer 42

I'm confused because its still -18

Answer 43

a is -18 yes
but what is f(3)

Answer 44

-6

Answer 45

right?

Answer 46

we found a to be -18
and our function was f(x)=-2x+18+a
or f(x)=4x+a
yes f(3)-6

Answer 47

f(3)=-6*

Answer 48

it was wrong :/

Answer 49

how

Answer 50

idk :/

Answer 51

y=4x+a and y=-2x+b
set them equal as x goes to 3
12+a=-6+b
18+a=b
so
we have
y=-2x+(18+a)
f(0)=0
so
0=0+18+a
a=-18
so we have
y=4x-18
f(3)=4(3)-18=12-18=-6

Answer 52

@wio did i do something wrong here?

Answer 53

I'm not sure, but you did over complicate it a bit.

Answer 54

\[
\int_0^xf'(t)~dt = f(x)-f(0) = f(x)
\]

Answer 55

So \[
\int_{0}^af'(x)~dx = f(a)
\]

Answer 56

For this problem though, we are just finding the area under the curve.

Answer 57

For example: \[
f(2) = 2\times 2=4
\]

Answer 58

can we integrate f' to t=0 to t=3
f' isn't continuous on [0,3]

Answer 59

\[
f(3)-f(2) = -2\times 1 = -2\implies f(3) = -2+f(2) = -2+4 =2
\]

Answer 60

In general, when we have a single discontinuity \(c\) in the interval \((a,b)\), then we define:\[
\int_a^bg(x)~dx = \lim_{t\to c^+}\int_a^tg(x)~dx+\lim_{u\to c^-}\int_u^bg(x)~dx
\]

Answer 61

I guess I just don't understand how the way I mentioned led to a wrong answer

Answer 62

i integrated both sides of x=3
the integrated function we suppose to be continuous on both left and right side
so that means the right limit=left limit as x approaches 3
then used f(0)=0

Answer 63

Okay, can you explain to me again what you did?

Answer 64

so we wanted to evaluate f(3)
I was trying to find f(3) given f(3 to right)=4x+a
f(3 to left)=-2x+b
f is suppose to be continuous so I want to find when f(3 to right)=f(3 to left)
4(3)+a=-2(3)+b
18+a=b

so we can right f(x)=-2x+18+a
f(0)=0 so f(0)=18+a=0
which means a=-18

Then y=-2x or we can say y=4x-18

this does after all give us f is continuous at x=3
since f(3)=-2(3)=-6 and f(3)=4(3)-18=-6

and also we sill have y'=-2 to the left and y'=4 to the right

so I'm not seeing how it is wrong
what condition wasn't met

Answer 65

Don't you want to use \(2x+c_1\) and \(-2x+c_2\)?

Answer 66

well the picture says y'=4 to the right and y'=-2 to the left of x=3

Answer 67

First of all, we don't know that the anti derivative is continuous.

Answer 68

f is the antiderivative of f'
and is stated f is continuous

Answer 69

Ok, that is weird.