1. ## tangent line

I have 3 problems I'm struggling with. Any help would be great

1) Given the function below

f(x)=100x3+253
-Find the equation of the tangent line to the graph of the function at x=1. Answer in mx+b form.

(Now, I found the derivative (100x^2)/(100x^3+25)^2/3. I then plugged 1 in as x and got 4, and when putting it into the point-slope form equation, I get 4x. But for some reason, the test I'm taking is telling me this is wrong, so I'm stumped already.)

The rest goes..

-Use the tangent line to approximate f(1.1).
-Compute the actual value of f(1.1). What is the error between the function value and the linear approximation?
Answer as a positive value only. Approximate to 5 decimal places.

2. Let's explore the function below using the Mean Value Theorem.

f(x)=(7x-21)3There are two x-values where the slope of the tangent line to the graph equals the slope of the secant line between x=0 and x=9. To find them, follow the steps below.

[A] First, find the slope of the secant line between x=0 and x=9.
[B] Now, find the derivative of the function.
(I already know this one is 21(7x-21)^2)
C] Set the derivative equal to mSEC and solve. One of these values is x=0, what is the other?
(But with this one, I have no idea where x=0 comes from, and when I set it equal to the slope if get 23/7, which ends up not being correct).

3. box with a square base and open top must have a volume of 70304 cm3. We wish to find the dimensions of the box that minimize the amount of material used.

First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base.
[Hint: use the volume formula to express the height of the box in terms of x.]
Simplify your formula as much as possible.
(With this one, I need A(x), A'(x), A"(x), and the evaluation of A"(x) at the x value where A'(x) = 0)

Pleeeasse help. If you can help with any, I would be in your debt for sure!

2. ## Re:

Originally Posted by tyuolio
I have 3 problems I'm struggling with. Any help would be great

1) Given the function below

f(x)=100x3+253
-Find the equation of the tangent line to the graph of the function at x=1. Answer in mx+b form.

(Now, I found the derivative (100x^2)/(100x^3+25)^2/3. I then plugged 1 in as x and got 4, and when putting it into the point-slope form equation, I get 4x. But for some reason, the test I'm taking is telling me this is wrong, so I'm stumped already.)

The rest goes..

-Use the tangent line to approximate f(1.1).
-Compute the actual value of f(1.1). What is the error between the function value and the linear approximation?
Answer as a positive value only. Approximate to 5 decimal places.
RE:

3. ## Re:

Originally Posted by tyuolio
I have 3 problems I'm struggling with. Any help would be great

First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base.
[Hint: use the volume formula to express the height of the box in terms of x.]
Simplify your formula as much as possible.
(With this one, I need A(x), A'(x), A"(x), and the evaluation of A"(x) at the x value where A'(x) = 0)

Pleeeasse help. If you can help with any, I would be in your debt for sure!
RE:

4. #2c
21(7x-21)^2=9261
(21)[7(x-3)]^2=9261
49(x-3)^2=9261/21
49(x^2-6x+9)=441
x^2-6x+9=441/49
x^2-6x+9=9
x^2-6x=0
x(x-6)=0
x=0 or x=6

5. actually everybody, the first problem's equation is (100x^3+25)^1/3, sorry can you all still help?

6. You solved corectly for the slope, m=4.
Going back to the initial function you have to find the
f(1)=(100(1)^3+25)^1/3-125^1/3=5
y-5=4(x-1)
y=4x-4+5
y=4x+1

7. see the red explanations.
Originally Posted by tyuolio
I have 3 problems I'm struggling with. Any help would be great

1) Given the function below

f(x)=100x3+253
-Find the equation of the tangent line to the graph of the function at x=1. Answer in mx+b form.

(Now, I found the derivative (100x^2)/(100x^3+25)^2/3. I then plugged 1 in as x and got 4, and when putting it into the point-slope form equation, I get 4x. But for some reason, the test I'm taking is telling me this is wrong, so I'm stumped already.)

y=4x+1

-Use the tangent line to approximate f(1.1).
fappx(1.1)=4(1.1)+1=5.4, use the tangent equation, put x=1.1
-Compute the actual value of f(1.1).
f(1.1)=[100(1.1)^3+25]^1/3=5.40726
What is the error between the function value and the linear approximation?
Answer as a positive value only. Approximate to 5 decimal places.
f(1.1)-fappx(1.1)=0.00726

8. Originally Posted by qbkr21
RE:
would A"(x) be 2+562432/x^3 then? and if you evaluate the x value from before with this equation, would it be the same as how you calculated it with the h value that it never says to find?

9. ## Re:

Re:

10. Originally Posted by qbkr21
Re:
I don't know. It says "We next have to make sure that this value of x gives a minimum value for the surface area. Let's use the second derivative test. Find A"x."

11. ## Re:

What do you mean by evaluate x. There is no x to evaluate...

-qbkr21

12. Originally Posted by qbkr21
What do you mean by evaluate x. There is no x to evaluate...

-qbkr21
Now, calculate when the derivative equals zero, that is, when ${A}'{\left({x}\right)}={0}$.
According to you, this x value was 52 (you used the variable "w" insted).
So I guess you take this x (w) value and input it into the A"(x) function?

13. ## Re:

Yes; along with h (height) that you found by plugging in w (width) equation. Plug in both of these found values into the original Area equation to find Max or Min Surface Area.

-qbkr21

14. Originally Posted by qbkr21
Yes; along with h (height) that you found by plugging in w (width) equation. Plug in both of these found values into the original Area equation to find Max or Min Surface Area.

-qbkr21
qbkr21 found the area, but he did not make sure that this is the minimum area.
tyuolio, you are right.
You have to use the second derivative test to make sure that the area found is the one you are looking for.
You have to show that A''(52)>0 to finalize you problem.
Since A''(52)=2+562432/52^3=2+4=6>0.
You have a minimum area for x=52