Question

How do you find the equation of the line tangent to `f(x)=3/x^2` at x=2?

Answer 1

`y=-3/4x+9/4`

Explanation:

First we can find the point that the tangent line will pass through, which is the point on `f` at `x=2`.

`f(x)=3/2^2=3/4`

The tangent line passes through the point `(2,3/4)`.

Next, we need to find the slope of the tangent line. This is the definition of the derivative, which gives the slope of the tangent line at a given point.

To differentiate the function, we will first need to write the function:

`f(x)=3/x^2=3x^-2`

Then, we will use the power rule to find the derivative. The power rule states that if `f(x)=x^n`, then `f'(x)=nx^(n-1)`.

Constants that the function are multiplied by stay being multiplied by the function, and are not altered. This can be viewed alongside the power rule as if `f(x)=ax^n`, then `f'(x)=anx^(n-1)`.

So, we see that where `f(x)=3x^-2`:

`f'(x)=3(-2)(x^(-2-1))=-6x^-3=-6/x^3`

The slope of the tangent line at `x=2` is:

`f'(2)=-6/2^3=-6/8=-3/4`

So, the tangent line has slope `-3/4` and passes through the point `(2,3/4)`.

A line that passes through the point `(x_1,y_1)` and has slope `m` can be written as:

`y-y_1=m(x-x_1)`

So, with the given information, this becomes:

`y-3/4=-3/4(x-2)`

Simplified, this becomes:

`y=-3/4x+9/4`

We can graph `f` and the tangent line:

graph{(y-3/x^2)(y+3/4x-9/4)=0 [-3.8, 7.296, -1.983, 3.564]}

The line is tangent at `x=2`.