Tangent line to f(x) = x^3 - 2x at x = -1 Step by Step

Step by step

The question

Question: find the tangent line to f(x) = x³ − 2x at x = −1.

f(x) = x³ − 2x

1 Find the point

The line touches the curve at x = −1.

Its y is f(−1): put −1 in.

We get 1.

f(−1) = (−1)³ − 2(−1) = −1 + 2 = 1

So the point is (−1, 1).

point = (−1, 1)

2 Find the slope

The slope of the curve is the derivative.

Use the power rule on each term.

f′(x) = 3x^(3 − 1) − 2 × 1

Now we have f′(x) = 3x² − 2.

= 3x² − 2

At x = −1: put −1 in.

The slope is m = 1.

f′(−1) = 3(−1)² − 2 = 3 − 2 = 1

Careful!

The slope of a line is one number.

f′(x) changes with x, so we put x = −1 in.

Careful: not m = 3x² − 2 ✗

3 Rise = m × run

We call the slope m.

m = rise over run.

So the rise is m times the run.

rise = m × run

Like 8/2 is 4, so 8 is 4 times 2.

82 = 4 → 8 = 4 × 2

Take any point (x, y) on the line.

We go from (−1, 1) to (x, y).

The rise is y − 1: the change in y.

The run is x − (−1): the change in x.

rise = y − 1

run = x − (−1)

Put them in: rise = m × run becomes y − 1 = (x − (−1)).

y − 1 = (x − (−1))

Taking away a negative is the same as adding.

− (−) = +

So it becomes y − 1 = (x + 1).

y − 1 = (x + 1)

This is called point-slope form.

It is made from a point and the slope.

m = 1, and 1 times a bracket is just the bracket.

y − y₁ = m(x − x₁)

4 Write as y = mx + b

A bracket with a plus or nothing in front can just be removed.

Now we have y − 1 = x + 1.

y − 1 = x + 1

We want y by itself.

The − 1 is in the way.

To remove − 1, we add 1.

We do it on both sides, so they stay equal.

y ~~− 1 + 1~~ = x + 1 + 1

− 1 and + 1 make zero.

Cross them out.

1 plus 1 is 2.

So y = x + 2.

y = x + 2

5 Check

Let's check on a picture.

The curve y = x³ − 2x.

Our line touches it at (−1, 1), with the same steepness.

Right!

Let's look back.

First: find the point.

Then: find the slope.

Then: rise = m × run.

Then: write as y = mx + b.

Last: check.

Answer