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

Answer 1

To answer this question, you'll need to solve #f'(x)=1#, which is:

#4x^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: #root(3)(1/8+sqrt(1/64+1/27))+ root(3)(1/8-sqrt(1/64+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)#

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Answer 2

Charles Anctil

The equation of the line tangent to the graph of f(x) = x^4 + 2x^2 at the point where f ' (x) = 1 is y = 4x + 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.