Question

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Let f(x)=x^4+2x^2, find the equation of the line tangent f where f'(x)=1

Answer 1

The slope of the tangent is 1. Find f'(x) and then set it equal to 1 to find the x-coordinate. Then you should be able to solve for the y and write the equation of the line

Answer 2

I can't solve for x when i set f'(x) equal to 1

Answer 3

what did you get for f'(x)?

Answer 4

4x^3+4x

Answer 5

There's not a nice way to solve it algebraically. You have to try it with a numerical method or a calculator

Answer 6

try Newton's method if you can't use a calculator

Answer 7

er, what?

f'(x) is the slope of the line

assuming you want to find the line tangent at x=1
then you can determine the coordinate point from f(1)= ...

f'(1) will give slope,
assuming you wnat a linear line, use point slope formula

Answer 8

but what are the coordinates for the point slope form? @sooobored

Answer 9

@mathmate

Answer 10

f(x)=x^4+2x^2
f(1)=1^4+2*1^2
f(1)=1+2
f(1)=3
(1,3)

Answer 11

the answer is y=x-0.122 but i want an explanation as to how to get this answer

Answer 12

oh, i think i misread the problem

Answer 13

find when the tangent line when f'(x)=1

we know f'(x)= 4x^3+4x

so we want to find x when
1=4x^3+4x

Answer 14

yeah

Answer 15

4x^3+4x-1=0
so find the cubic roots of x

Answer 16

That is where i am having trouble, I am unable to find x

Answer 17

fyi, its not a very nice problem, the root turns out to be
x~0.23673, other two roots are imaginary

Answer 18

so then we can plug in x in f(x) to find the y value, right?

Answer 19

yea

Answer 20

cubic formula
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Answer 21

Thankyou so much, I finally got the answer :D