Find Tangent Line To a Curve

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april19

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Sep 22, 2010
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What is the equation of the tangent line to the graph
f(x)=x^4+2x^2
at the point where f '(x)=1?

Since the derivative is 1, the slope of the tangent line is 1.
So for y=mx+b, I know m=1.
I have a hard time figuring out b (the y-intercept).

To figure out b, I think first I have to solve x after I find the derivative so I know where the tangent line touches the curve.

f '(x)=4x^3+4x=1
I tried looking for formulas to solve a polynomial with 3rd degree and it has some humungous formula and I got lost trying to solve it.

Please help me find this tangent line.
Thanks.
 
Hello, april19!

What is the equation of the tangent line to the graph
. . \(\displaystyle f(x)\,=\,x^4+2x^2\) at the point where \(\displaystyle f '(x)\,=\,1\,?\)
Click to expand...

Your work and your reasoning are both correct.
I don't think there is an elementary solution.

We must solve: .\(\displaystyle 4x^3 + 4x \:=\:1\) to determine \(\displaystyle P\), the point of tangency.

Suppose \(\displaystyle x = a.\)
. . then: .\(\displaystyle 4a^3 + 4a \,=\,1\) .[1]

Hence, the point \(\displaystyle P\) is: .\(\displaystyle (a,\:a^4+2a^2)\)

And the tangent is: .\(\displaystyle y - (a^4+2a^2) \;=\;1(x - a) \quad\Rightarrow\quad y \;=\;x + (a^4+2a^2-a)\) .[2]

But without the value of \(\displaystyle a\), we can go no further.


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I tried some surgery, but got nowhere.


From [1], we have:
. . \(\displaystyle 4a(a^2 + 1) \:=\:1 \quad\Rightarrow\quad a^2+1 \:=\:\frac{1}{4a}\)

In [2], the y-intercept is:

. . \(\displaystyle a^2(a^2+2) - a \;\;=\;\;a^2(\underbrace{a^2+1}_{\text{This is }\frac{1}{4a}} +1) - a \;\;=\;\; a^2\left(\tfrac{1}{4a} + 1\right) - a\)
. . . . . . . . . . . . .\(\displaystyle =\;\;\frac{a}{4} + a^2 - a \;\;=\;\;a^2 - \frac{3a}{4} \;\;=\;\;\frac{a(4a-3)}{4}\)

And we still need the value of \(\displaystyle a.\)

 
I also tried what Soroban attempted, but got no where as well.

Without resorting to cumbersome formulas, run it through a solver on a calculator.

\(\displaystyle 4x^{3}+4x=1\)

\(\displaystyle x=.236732903865\)

Now, you can finish.

Or, if you must do it algebraically, try Newton's method (assuming you have learned that at this point).

Use \(\displaystyle x_{n+1}=x-\frac{4x^{3}+4x-1}{12x^{2}+4}\)

Use an initial guess of, say, 0.5.

\(\displaystyle .5-\frac{4(.5)^{3}+4(.5)-1}{12(.5)^{2}+4}=.285714285714\)

Now, place this result back into the formula.

Perform the interations and it will converge to x=.236732903865
 
Wow. Thank you very much. I knew it was not easy but didn't think it's that hard.
This problem is supposed to be for a high school Calculus class so how in the world can a high schooler solve that.
Anyway, thanks a bunch for all your help.
 
Just as a check, are you sure it did not say at f(x)=1 instead of f'(x)=1?.
 
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