Tangent line to y = 2x^3 - x at x = 1 Step by Step

Step by step

The question

Question: find the tangent line to y = 2x³ − x at x = 1.

y = 2x³ − x

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) = 2(1)³ − 1 = 2 − 1 = 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) = 2 × 3x^(3 − 1) − 1

Now we have f′(x) = 6x² − 1.

= 6x² − 1

At x = 1: put 1 in.

The slope is m = 5.

f′(1) = 6(1)² − 1 = 6 − 1 = 5

Careful!

The slope of a line is one number.

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

Careful: not m = 6x² − 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, 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 = 5(x − 1).

y − 1 = 5(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 5 in front of the bracket multiplies every part inside.

So we write 5 times each part.

y − 1 = 5 × x + 5 × (−1)

Work out each multiplication.

Now we have y − 1 = 5x − 5.

y − 1 = 5x − 5

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~~ = 5x − 5 + 1

− 1 and + 1 make zero.

Cross them out.

−5 plus 1 is −4.

So y = 5x − 4.

y = 5x − 4

5 Check

Let's check on a picture.

The curve y = 2x³ − x.

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