Question

Calculus Help?

Answer 1

@sourwing

Answer 2

just #22 please help

Answer 3

@ganeshie8 @Loser66

Answer 4

I am sorry, I do forget how to do.

Answer 5

@kc_kennylau

Answer 6

1) Calculate the slope of the blue line using the point of tangency and the x-intercept.
2) Using that slope, and the point of tangency, find the equation of the blue line.
3) Use the linear approximation just created to calculate the desired value.

Answer 7

Let me try, hehehe. Hopefully I am not wrong,
m = -1 --> tangent line for 22) is y -1=-(x-2)--> y = -x +3
at x =1.9 ---> y = -1.9 +3 = 1.1
ha!! it looks weird

Answer 8

@tkhunny correct me, please

Answer 9

because that is the tangent line, not the curve f.

Answer 10

I'm only concerned that we have not "used differentials". Using a differential is equivalent to traveling the tangent line, but it does have a little different flavor.

If we have y = f(x), then dy = f'(x)dx

To find the approximation to f(1.9), we have x = 2 and dx = -0.1.
We also have y = 1 = f(x), x = 2, and f'(2) = -1

dy = (-1)(-0.1), giving dy = 0.1 and f(1.9) = f(2 - 0.1) = 1 + 0.01 = 1.1

Same thing as just traveling the tangent line. This is also sometimes called the "linearization".

Answer 11

@tkhunny I understand a little what you posted. So now I have to do f(2.04)?

Answer 12

@ganeshie8 @mathmale please help? I don't fully understand

Answer 13

It is almost exactly the same.

x = 2
dx = 0.04

Answer 14

ok. thanks

Answer 15

Hello, VB,
Let's take a look at Problem #22.
First of all, let's begin with the formula for the linearization L(x) of a given function at a given point on the graph:
\[L(x)=f(x _{0})+f'(x _{0})(x-x _{0).}\]
Here, that f(x) is not specifically defined as a function, but it is defined as a graph. We are interested in finding a linearization of this function at the point (2,1), which is shown on the graph. Therefore, when x=2, y=1. Or, in other words, when x=2, f(2)=1.

Therefore, \[f(x _{0})=f(2)=1.\]

Answer 16

Look carefully at that blue tangent line. What's its slope? Extend that blue line through the y-axis at (0,3). What's the rise? What's the run? What's the slope of that line? (Answer: -1.)

Therefore, \[f'(x _{0})=-1.\]

Answer 17

Now put all of this info together:
The linearization of the function at (2,1) is\[L(x)=f(x _{0}+f'(x _{0})(x-x _{0})=1-1(x-2).\]

You could either leave it like this, or simplify it to L(x)=1-x+2=-x+3.

Answer 18

Now check this result: Suppose x=2, we expect that y=1.
Substitute x=2 into L(x)=-x+3.

Answer 19

Result is 1 (as expected).

Answer 20

Therefore, the desired linearization is L(x)=3-x.