What is the equation of the line tangent to the

What is the equation of the line tangent to the graph of $f\left(x\right)={x}^{4}+2{x}^{2}$ at the point where f'(x)= 1?
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Sterling Mcdaniel

To answer this question, you'll need to solve f'(x)=1,
which is:
$4{x}^{3}=4x=1.$
This equation has no rational solution, so you'll need to approximate, by some method either technological or otherwise. (Or use the general formula for the solution of a cubic to get:
$\sqrt[3]{\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{1}{27}}}+\sqrt[3]{\frac{1}{8}-\sqrt{\frac{1}{64}+\frac{1}{27}}}$
Once you get an approximation, call it a, find f(a) to get the point: (a, f(a)) on the curve. And, since f'(a)=1, we get:
The equation of the tangent line is:
y=x-a+f(a)