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| Math Central | Quandaries & Queries |
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Question from Princess, a student: Hi! I'm a college sophomore student and I am taking a Business Calculus class. And I'm having a REALLY hard time trying to figure out this problem: If f ' (x) = 3x^2 +1, find the equation of the tangent line to f(x) = x^3 + x at x= -1. Thanks! |
Hi Princess.
The term f ′ (x) refers to the derivative of f(x). The derivative of f(x) is the slope of f(x). The slope of f(x) is the tangent line to f(x) at whatever value of x you are interested in.
So f ′ (x) is itself the slope of f(x). This means 3x2 + 1 is the slope of x3 + x.
By simply plugging in the value x = -1, you can find the slope of the tangent line. And of course f(x) is the y value corresponding to x, giving you a point on the tangent line (the point of tangency itself).
The equation of any line, given a point on the line (x0, y0) and a slope m is:
y - y0 = m (x - x0)
Since the slope of a function f(x) is f ' (x) and the value y0 = f(x0), the tangent line of any function f(x) at a particular value of x = x0 is:
y - f(x0) = f '(x0) (x - x0)
Just substitute x0 = -1 and simplify to complete the question:
y - f(-1) = f '(-1) (x + 1)
y - ((-1)3 + (-1)) = (3(-1)2 + 1)(x + 1)
y - (-2) = (4)(x + 1)
y = 4x + 2.
Cheers,
Stephen La Rocque.
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