Question

Calculus help?

Answer 1

I have an extra credit assignment due tomorrow because I bombed my final, so I really need help!

Answer 2

Could you post your question?

Answer 3

Use the following graph of f (x) to find the graph of f ‘(x).

Answer 4

141 is the graph of f(x) which one would be the graph of f'(x)?

Answer 5

Notice how this point is the min (see attached)

Answer 6

ok

Answer 7

At this point, the slope of the tangent line is 0. So that means that the graph of the derivative will have a root slightly above x = 4

Answer 8

Looking at the answer choices, that means we can eliminate graph 142 and 144 because they have roots below x = 4

Answer 9

okay

Answer 10

So it's between 143 and 145

I'm not 100% sure on this, but it looks like graph 145 has the root in a slightly more accurate location than graph 143 (since it's more to the right a bit). So I'm leaning towards 145.

Answer 11

I have a few more questions on the differentiability section if you can help?

Answer 12

sure I can do a few more

Answer 13

If v(t) = 2t2 + t, the velocity of an object over time, what is the acceleration of the object at t = 0?

Answer 14

Acceleration is the derivative of velocity (since acceleration is change in velocity over change in time)

ie...

derivative of velocity = acceleration

Answer 15

v(t) = 2t^2 + t

d/dt[v(t)] = d/dt[2t^2 + t] ... derive both sides with respect to t

v ' (t) = 4t + 1

a(t) = 4t + 1 ... this is the acceleration function

Now plug in t = 0

a(t) = 4t + 1

a(0) = 4(0) + 1

a(0) = 1

So the acceleration is 1 unit per second per second (replace 'unit' with feet, meters, or whatever unit you're using)

Answer 16

A differentiable function has the value y(2) = -1 and the derivative value y′(2) = 3. Approximate the value of y(1.9).

Answer 17

hint: the slope of the tangent line of this function at x = 2 is m = 3

the point (2,-1) lies on the function

Answer 18

your goal is to find the equation of the tangent line (with slope 3 and going through the point (2,-1))

once you have this function f(x), you can use it to find f(1.9)

Answer 19

-1.3 is my answer.

Answer 20

If position is given by p(t) = t^2 +1, find the velocity v(t) at t = 2.

Answer 21

y - y1 = m(x-x1)

y - (-1) = 3(x - 2)

y + 1 = 3x - 6

y = 3x - 6 - 1

y = 3x - 7 ... equation of the tangent line at (2,-1)

y(1.9) = 3(1.9) - 7

y(1.9) = -1.3

so you are correct, nice work

Answer 22

p(t) = t^2 +1

p ' (t) = 2t .. derive both sides

now plug in t = 2

p ' (2) = 2*2

p ' (2) = 4

the velocity is 4 units per second (replace 'units' and 'second' with the appropriate units if need be)

Answer 23

I have 3 more graph questions in the derivative section of my EC.

Answer 24

sorry I can't open docx files (my computer is a bit old)

Answer 25

thanks

Answer 26

np:)

Answer 27

In the first one, notice how the graph has 2 local extrema. That must mean that there are 2 roots in the derivative. The only one that fits is the third choice (the parabola)

Answer 28

i had a thought that it was that one but i didn't know why

Answer 29

what did you get for the next one?

Answer 30

hint: the given graph is the graph of the slopes (since f ' (x) represents the slope of the tangent line)

Answer 31

so its a or be right?

Answer 32

for choice A or B, is the graph increasing or decreasing when x < 0

Answer 33

decreasing

Answer 34

you sure?

Answer 35

no

Answer 36

notice how the graph goes up as you move from right to left (when x < 0)

Answer 37

so the graph is increasing on the interval (-infinity, 0)

but hold on a second...look at the graph of f ' (x), notice how the portion to the left of the y axis is below the x-axis, so you have negative slopes which imply that the graph should be decreasing

Answer 38

so it would be c or d.

Answer 39

which one are you thinking?

Answer 40

and if you can, state why you think that's the answer

Answer 41

i am not sure