Find the slope of the tangent line to the graph of the...

Find the slope of the tangent line to the graph of the function at the given point, and determine an equation of the tangent line.
$$f(x)=-\frac{1}{x} \operatorname{at}\left(3,-\frac{1}{3}\right)$$

Key Concepts

Point-Slope Form Equation

The point-slope form is a method for writing the equation of a line when a point on the line and its slope are known. Expressed as y - y? = m(x - x?), it is particularly useful in constructing the equation of the tangent line once the slope (from the derivative) and a point of tangency have been identified.

Tangent Line

The tangent line to a function at a specific point is the straight line that touches the function at that point and has the same slope as the function. This line serves as a local linear approximation to the function and its slope is determined by the derivative at that point.

Derivative

A derivative represents the instantaneous rate of change of a function with respect to its variable. It quantifies how the function value changes as the input changes and is fundamentally used to find the slope of the tangent line to the graph at a given point, providing insight into the behavior of the function.

Transcript

00:01 We're doing another tangent line.

00:04 And if you're doing a tangent line, you need the point, which they gave us, is the ordered pair 3, negative 1 3.

00:14 And the other thing that we need to find is this slope.

00:17 When the slope is found by finding the derivative and plugging in the x coordinate 3 into the problem.

00:25 So when they are looking at f of x is equal to negative 1 over x.

00:32 The first step to finding the derivative would be to figure out what f of x plus h is, which is just replacing the x in the problem with x plus h.

00:42 So if your next step is to do f of x plus h minus f of x, then it makes sense to just rewrite negative 1 over x plus h.

00:53 And instead of subtracting the negative, let's switch that to plus 1 over x.

00:57 And then we can simplify that by getting the same denominators.

01:01 So i get the same denominers by multiplying x and x plus h by each other.

01:07 So the left fraction needs to be multiplied by x, and then the right fraction needs to be multiplied by x and h, x plus h.

01:17 And what happens right there is that the x is canceled.

01:22 So what are we left with is to find the derivative.

01:27 It's equal to the limit as h approaches zero of what we just found that h over, h over x times x plus h, all over h.

01:42 Well, instead of dividing everything by h, i'm going to multiply by the reciprocal.

01:47 And i hope if you're in calculus, you understand that dividing by h is the same thing as multiplied by 1 over h...

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