# Let f(x) = \frac{2}{x^2}. a. Find the slope of the tangent line to the graph of this function at...

## Question:

Let {eq}\displaystyle f(x) = \frac{2}{x^2} {/eq}.

a. Find the slope of the tangent line to the graph of this function at {eq}x = 3 {/eq}.

b. Find the equation of the tangent to the graph of this function at {eq}x = 3 {/eq}.

## Tangent Line

We can use the first derivative to find the instantaneous rate of change of a function. Since the rate of change is a slope, we can use it to construct the line that has the same instantaneous slope.

a. In order to find the slope of the tangent line, we need to take the first derivative. We can rewrite our function using a negative exponent so that we can easily use the Power Rule.

{eq}f(x) = 2x^{-2}\\ f'(x) = -4x^{-3} {/eq}

By evaluating this at x = 3, we will find the slope of the tangent line to that point.

{eq}f'(3) = -4(3)^{-3} = -\frac{4}{27} {/eq}

b. In order to use this slope to construct the tangent line, we need both the x and y coordinate of the point on the original function where we want this line to touch. We have the x value, so we can evaluate our function at this point to find the y value.

{eq}f(3) = \frac{2}{(3)^2} = \frac{2}{9} {/eq}

Now that we have both a point and a slope, we can construct the tangent line using point-slope form.

{eq}y - \frac{2}{9} = -\frac{4}{27} (x - 3)\\ y = -\frac{4}{27} x + \frac{4}{9} + \frac{2}{9}\\ y = -\frac{4}{27} x + \frac{2}{3} {/eq}

Thus, the tangent line to this function at the given point is {eq}y = -\frac{4}{27} x + \frac{2}{3} {/eq}.