Question

For `f(x) =-x^-3+x^-2`, what is the equation of the line tangent to `x =2`?

Answer 1

Here's two ways to do it. I get

`f_T(2) = 7/16 x - 3/4`

graph{(y - (1/4 - 1/16x))(y - (-x^(-3) + x^(-2))) = 0 [-1.127, 5.032, -1.836, 1.243]}

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LINEARIZATION METHOD

The easy way to do it is via this equation:

`bb(f_T(a) = f(a) + f"'"(a)(x+a))`

where `f_T` is the function for the tangent line, `f` is the function, and `f'` is its derivative. `a` is the `x` value at which you are evaluating `f_T`.

When you take the derivative of polynomials, use the power rule:

`d/(dx)[x^n] = nx^(n-1)`

Therefore:

`f'(x) = -(-3)x^(-4) + (-2)x^(-3)`

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

And so at `a = 2`,

`color(green)(f'(2)) = 3/(2^4) - 2/(2^3)`

`= 3/16 - 2/8`

`= 3/16 - 4/16 = color(green)(-1/16)`

And finally,

`color(green)(f(2)) = -(2)^(-3) + (2)^(-2)`

`= -1/8 + 1/4`

`= -1/8 + 2/8 = color(green)(1/8)`

So, using the first equation, we then get:

`color(blue)(f_T(2)) = f(2) + f'(2)(x+2)`

`= 1/8 - 1/16(x+2)`

`= 2/16 - 1/16 x + 2/16`

`= color(blue)(-1/16 x + 1/4)`

REGULAR WAY

Or, you could take the derivative, evaluate it at `x = 2`, then extrapolate to the `y`-intercept. It's a bit harder but does not require that you memorize much of an equation, other than slope is `(Deltay)/(Deltax)`.

`color(green)(f'(2) = -1/16)`, from before.

So, you have the slope. Your equation is so far:

`f_T(2) = -1/16x + ???`

Then, plug in `f(2)` and use the slope formula to find the `y`-intercept.

`color(green)(f(2) = 1/8)`, from before.

Therefore:

`(Deltay)/(Deltax) = (1/8 - y_1)/(2 - 0) = -1/16`

Now, solving for `y_1`:

`-1/8 = 1/8 - y_1`

`color(green)(y_1) = 1/8 + 1/8 = color(green)(1/4)`

So the end result is the same:

`color(blue)(f_T(2) = -1/16 x + 1/4)`

This is just more conceptual/visual.