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

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Since the tangent line to function f(x) = x ^3 has to be parallel to the line 3x - y + 1 = 0, both needs to have the same slope.
Equation of the straight line can be rewritten in slope-intercept form to : y = 3x - 1. Hence the desired slope of our line is 3.

And derivative of f(x) will provide the slope of the line tangent to f(x)
d/dx(f(x)) = 3x
setting it equal to 3
3x = 3 or x = 1 provides one of the conditions

Corresponding y coordinate can be found from the f(x) = x^3 equation, y = (1)^3 = 1
The slope of the desired line is equal to 3, and we now have a point on the desired line (1,1), as...

(The entire section contains 2 answers and 351 words.)

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