Find Tangent Line To a Curve

[april19]

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.

 

[soroban]

Hello, april19!

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

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 $\displaystyle a.$

 

[galactus]

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

 

[april19]

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.

 

[galactus]

Just as a check, are you sure it did not say at f(x)=1 instead of f'(x)=1?.

 

[april19]

I just double checked. It said f ' (x)=1.