Integration given slope of tangent at specific point

[B18]

Homework Statement

Find f if f"(x)=12x²+2 for which the slope of the tangent line to its graph at (1,1) is 3.

Homework Equations

The Attempt at a Solution

What I did first was found f(x)=x⁴+x²+cx+d (cx and d being constants of integration.) and from this point I attempted solving for cx and d but was stuck here. It may just be the wording of the problem but I'm not sure where else to go. How do I know what to set f(x) equal to in order to solve cx and d?

Thank you for any suggestions/help.

 

[Michael Redei]

"The tangent to the graph of f through the point (α,β) is τ" means:

(1) The graph of f passes through this point (α,β).
(2) The value of f'(x) for x=α is τ.

This gives you two equations, f(...)=... and f'(...)=..., and that's all you need to find the two unknowns, c and d in your case.

 

[B18]

So once I find c and d I plug them into f(x). Which is my final answer. I managed to find c= -3 by solving f'(1)=3. When I go to solve d what am I setting f(x) equal to? I found f(x) to be x^4+x^2-3x+d. I set this equal to 3 and plugged in 1 for x and got d= 4 making my answer f(x)=x^4+x^2-3x+4 however that equal does not pass through the point (1,1)?

 

[Dick]

So once I find c and d I plug them into f(x). Which is my final answer. I managed to find c= -3 by solving f'(1)=3. When I go to solve d what am I setting f(x) equal to? I found f(x) to be x^4+x^2-3x+d. I set this equal to 3 and plugged in 1 for x and got d= 4 making my answer f(x)=x^4+x^2-3x+4 however that equal does not pass through the point (1,1)?

If you want your curve to pass through (1,1) you want to set f(1)=1. Not f(1)=3.

 

[B18]

Ok my final answer is f(x)=x₄+x₂-3x+2. Does that seem correct?

 

[Michael Redei]

Ok my final answer is f(x)=x₄+x₂-3x+2. Does that seem correct?

You can check that quite easily by calculating f(1) and f'(1), which should be 1 and 3, respectively. (They are, so your solution is correct.)

 

[B18]

Wonderful, thank you for the help guys.

 

[Mark44]

Ok my final answer is f(x)=x₄+x₂-3x+2. Does that seem correct?

Make that f(x)=x⁴+x²-3x+2.

To add to what Michael said, it's easy to check, and something you should get into the habit of doing. You've already done all the heavy lifting.

There are basically three things you need to check:
1. Is f''(x) = 12x² + 2?
2. Is f(1) = 1?
3. Is f'(1) = 3?

If you can answer yes to all three, then your solution is correct.

 

[B18]

Thank you mark. I just had a hard time remember what statements meant what. If that makes sense. I did do the check and they all worked out thanks again.