Parallel tangent line

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raidershero

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I need help with this problem on my review sheet for Calc 1.

At how many values of 'x' is the function f(x) = (4x2-5x+1)(x-3) parallel to y=3x - 9?

What I've done is take the derivative of f(x) and I end up with:
f'(x) = 12x2 - 24x + 16

In order to solve this problem aren't I supposed to set this equal to the slope of y, which is 3?
If this is indeed the case then I would need to use the quadratic formula. Doing so, however, results in a negative number under the radical. What am I doing wrong? The answer key says the answer should be 2.

Thanks.
 
Hmmm -- I get two values for x, neither of which is 2.

Will you please verify that you correctly posted f(x) ?

Cheers
 
mmm4444bot said:
Double-check your arithmetic; the value in red (above) is not correct. :cool:
Click to expand...

I'll take you through what I have step by step so as to avoid any confusion.

Using the product rule I set my problem up:

f '(x) = (4x2 - 5x +1)*[x-3]' + (x-3)*[4x2 - 5x +1]'

This gives me:

f '(x) = (4x2 - 5x +1)*(1) + (x-3)*(8x - 5)

f '(x) = 4x2 - 5x +1 + 8x2 - 5x - 24x + 15 (I see that I had left out the 5x in red now)

**edit**
f '(x) = 12x2 - 34x + 16

Put this through the quadratic formula and it's just ugly. I must be missing something else here.
 
Last edited:
mmm4444bot said:
Hmmm -- I get two values for x, neither of which is 2.

Will you please verify that you correctly posted f(x) ?

Cheers
Click to expand...
Yep. That's exactly as it appears on the review handout. I'm getting something not even remotely close to 2. I get 78 + or - something that calculates out to approximately -4.48.

**EDIT**
Haha I take that back. I don't know how I got that. I'm getting 34 + or minus 2*sqrt(97)/24.

And just to clarify, the solution is not '2'. There are '2' values where the functions are parallel. Do I have it right now?
 
Last edited:
f '(x) gives the slope of f(x) at the value x.
Using the quadratic formula gives two results of 2.37 and 0.45.
At these two values of x, the slope of f(x) is 3.
These are the two value of x at which f(x) is parallel to y=3x-9.
So the answer is 2.
 
KindofSlow said:
f '(x) gives the slope of f(x) at the value x.
Using the quadratic formula gives two results of 2.37 and 0.45.
At these two values of x, the slope of f(x) is 3.
These are the two value of x at which f(x) is parallel to y=3x-9.
So the answer is 2.
Click to expand...

Those aren't the numbers I'm getting though. How is it you are coming up with these values for x?

The decimal values I'm getting are 34.82 and -33.18
 
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You want to find out where the slope of f(x) = 3.
f '(x) tells you the slope of f(x).
So set f '(x) = 3.
Now to use the quadratic formula, your quadratic equation must = 0 so don't forget to subtract 3 from both sides.
Now use the quadratic formula to solve for x.
 
KindofSlow said:
You want to find out where the slope of f(x) = 3.
f '(x) tells you the slope of f(x).
So set f '(x) = 3.
Now to use the quadratic formula, your quadratic equation must = 0 so don't forget to subtract 3 from both sides.
Now use the quadratic formula to solve for x.
Click to expand...

OK. That's basically the same thing I just did only I forgot to set it to zero first. It only changes my answer slightly though. Something tells me I screwed up the derivative. I have an enormous 34 hanging out in front of the radical in my quadratic formula.

No matter how many times I go over it I can't figure out where I took a wrong turn though.:confused:
 
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I figured out what I was doing wrong...:roll:

I wasn't including the "-b" in the numerator. It's been a very long time since I've had to use the quadratic formula. My answers are coming out very similar now. I have 2.38 and -.51. Still not quite sure why they're off slightly though. Thanks for all the help.:-D
 
KindofSlow said:
So the answer is 2.
Click to expand...

Ah, yes, I lost track of the original question. (Thanks, KindofSlow).

We don't need to find the actual zeros of f`(x). Simply checking the Discriminant would have confirmed two solutions.
 
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