If f is the function whose graph is shown below, let h(x) = f(f(x)) and g(x) = f(x2). use the graph of f to estimate the value of each derivative.

We want to estimate the value of two derivates by knowing the graph of the function f(x) and the derivation rules.

The estimates are:

g'(2) = 4

h'(2) = 0.5

We know that the derivation of a composition:

k(x)  = f(g(x))

is given by:

k'(x) = f'(g(x))*g'(x).

Now we have:

f(x) is the function shown in the graph.

h(x) = f(f(x))

g(x) = f(x^2)

Now we want to get both of these derivated and evaluated in x = 2.

We will get:

h'(x) = f'(f(x))*f'(x)

Evaluating that in x = 2 we get:

h'(2) = f'(f(2))*f'(2)

By looking at the graph we estimate:

f(2) = 1

f'(2) = -1

h'(2) = f'(1)*-1

The derivate at x = 1 is something near -0.5, then we got:

h'(2) = -0.5*-1 = 0.5

For g(x) we have:

g'(x) = f'(x^2)*(2*x)

Then:

g'(2) = f'(2^2)*(2*2)

       = f'(4)*4 = 1*4 = 4

If you want to learn more, you can read:

link

1. For the function h(x)=f(f(x)):

Use the chain rule to determine the derivative - h'(x):

From the graph f(2)=1, f'(f(2))=f'(1)=-1 (the tangent line to f(x) at x=1 has a slope -1) and f'(2)=-1 (the tangent line to f(x) at x=2 has a slope -1). Therefore,

2. For the function

Use the chain rule to determine the derivative - g'(x):

From the graph f'(4)=1 (the tangent line to f(x) at x=4 has a slope 1) and g'(2)=4.

h'(2)=1, g'(2)=4.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.