calc question?

Question

find an equation of the line that is tangent to the graph of F and parallel to the given line:
f(x) = x^3 , 3x - y + 1 = 0.

Answer 1

The tangent line is simply the derivative of f(x). To find a tangent of a specified slope, you need to identify the point on the graph which has that slope. Now, let's do it:
f'(x) = 3x^2. The slope of the given line is dy/dx where y = 3x+1, or 3. Hence, we want f'(x) = 3x^2 = 3, or x = +/- 1. At these values of x, y = +/- 1. So we need the equation of a line that passes through (1, 1) and has a slope of 3. You take it from here.

Answer 2

well, you can get the equation for the slopes of the lines tangent to f(x)=x^3 by taking the derivitive of the function

f'(x)=3x^2

writing the parallel straight line in slope intercept form gives:
y=3x+1

to be parallel with the straight line, the tangent line has to have the same slope, so we want the place where the 3x^2=3

which is where x=1 or x=-1

the question says find AN equation so we don't have to find them all

f(1)=(1)^3=1

so the place on the line f(x) that has the tangent line we are looking for is at point (1,1)

y=mx+b

we know m=3 and we can calculate b

1=3(1)+b
-2=b

so the answer is y=3x-2

lets check it

the slope is 3 so it is clearly parallel to the given line

and, it goes through (1,1) as does f(x) so it clearly is a candidate line

is it TANGENT at (1,1)

f'(x)=3x^2
f'(1)=3

looks right to me

I did this very quickly so you better check my math and equations cause I could easily have screwed something up, but perhaps this will help

Answer 3

dy/dx = 3x^2
The slope of the given line is dy/dx or 3.
3 = 3x^2
=> x= +1.-1

At these values of y = +1,-1.
So we need the equation of a line that passes through (1, 1) and has a slope of 3. or from (-1,-1) and slope 3

:. y = 3x + c;
=> y = 3x -2 or y = 3x - 4.
Both these equation satisfy the criteria.
Hence, both constitute answer.