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

Step by step

The question

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

f(x) = x³ − x

1 Find the point

The line touches the curve at x = 1.

Its y is f(1): put 1 in.

We get 0.

f(1) = 1³ − 1 = 1 − 1 = 0

So the point is (1, 0).

point = (1, 0)

2 Find the slope

The slope of the curve is the derivative.

Use the power rule on each term.

f′(x) = 3x^(3 − 1) − 1

Now we have f′(x) = 3x² − 1.

= 3x² − 1

At x = 1: put 1 in.

The slope is m = 2.

f′(1) = 3(1)² − 1 = 3 − 1 = 2

Careful!

The slope of a line is one number.

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

Careful: not m = 3x² − 1 ✗

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, 0) to (x, y).

The rise is y − 0: the change in y.

The run is x − 1: the change in x.

rise = y − 0

run = x − 1

Put them in: rise = m × run becomes y − 0 = 2(x − 1).

y − 0 = 2(x − 1)

Taking away 0 changes nothing.

− 0: nothing changes

So it becomes y = 2(x − 1).

y = 2(x − 1)

This is called point-slope form.

It is made from a point and the slope.

y − y₁ = m(x − x₁)

4 Write as y = mx + b

The 2 in front of the bracket multiplies every part inside.

So we write 2 times each part.

y = 2 × x + 2 × (−1)

Work out each multiplication.

So y = 2x − 2.

y is already by itself.

y = 2x − 2

5 Check

Let's check on a picture.

The curve y = x³ − x.

Our line touches it at (1, 0), 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