Tangent Line Approximations (Again) Examples

Let f (x) = 4x² + 3.

(a) Find the tangent line to f at x = 3.
(b) Use the tangent line from part (a) to approximate f at 3.001.

(a) To find a tangent line we need a point and a slope. The point we want is

(3, f (3)) = (3, 4(3)² + 3) = (3, 39).

The slope is

f '(3) = 8(3) = 24.

Putting this together, we get the line

y = 24x – 33.

(b) We put x = 3.001 into the tangent line and see what we get:

24(3.001) – 33 = 72.024 – 33 = 39.024.

If f (1) = 2 and f ' (1) = 3, estimate f (1.2).

We're told the value and slope of f at 1 and asked to estimate f at 1.2. This means we want to draw the tangent line to f at x = 1, and find the value of that tangent line when x = 1.2.

In other words, we have y_old = 2 , slope = 3, and Δ x = 0.2.

We're asked to find y_new, which we know how to do.

We never had to find the equation for the tangent line! It was enough to know the slope of the tangent line, the value of y we started at, and the amount by which x changed.

The picture below shows the tangent line to the function f at x = 0. Estimate f (-0.1).

We have y_old = -2 and Δ x = -.1 (yes, Δ x is negative). We want y_new, which is the value of the tangent line when x = -0.1.

In order to find y_new we need to know the slope of the line. Since we're given two points on the line, we can figure that out.

The slope of the tangent line is

Then

When doing these problems, so long as Δ x is small we expect y_old and y_new to be pretty close together. After all, y_old and y_new are both on the same line, and if Δ x is small we're not moving very much on the line. In the previous example, we had y_old = -2 and y_new = -2.2. These are pretty close together, so the value y_new = -2.2 is believable. An answer like y_new = 20 would be a call to double-check our calculations.

The values y_new, y_old, and Δ x can also be used to find the slope of a tangent line to f. Don't think too hard about this! We're just revisiting the formula

The picture below shows a tangent line to f at x = a. What is f '(a)?

We don't even need to worry about which one is y_new and which one is y_old. We just look at the two points and find rise and run.

The slope of the line is

Since this line is tangent to f at a, the slope of the line must agree with the slope of f at a. We conclude

Let f (x) = cos² x. Use a tangent line approximation to estimate f (0.1).

We need to decide where to draw the tangent line. Since 0.1 is close to 0, x = 0 is a good bet. Then

y_old = f (0) = 1

and

Δ x = 0.1

The slope of the tangent line is f '(0). First let's find the derivative of f:

f '(x) = 2(cos x)(-sin x)

so

f '(0) = 0.

We get

Let f (x) = x⁵ + 3x⁴ – 2x² – x. Use a tangent line approximation to estimate f (0.9).

Let's draw the tangent line at x = 1, since 1 is a convenient number that's close to 0.9. Then we have

Δ x = -0.1

y_old = f (1) = 1 + 3 – 2 – 1 = 1

To find the slope we need to know f ':

f '(x) = 5x⁴ + 12x³ – 4x – 1

The slope is

f '(1) = 5 + 12 – 4 – 1 = 12.

This is enough to find the approximation:

It's kind of weird that we got a negative number, but if we look at a graph of the function and tangent line together we can see why: the tangent line drops below the x-axis while the function is still above the x-axis.

We know how to do tangent line approximations. We know how to do tangent line approximations by finding just the slope of the tangent line rather than bothering with the whole tangent line equation.

The only thing we've changed for these problems is that we're no longer telling you where to draw the tangent line. Instead, you get to figure that out for yourself. To pick the right spot for the tangent line, take x = a near where you're asked to estimate f, and make sure f (a) and f '(a) are easy to compute.

Can a tangent line approximation ever produce the exact value of the function? Why or why not?

Yes. If the function f is a straight line then the tangent line at any point will be the same as the function. This means the tangent line approximation will produce the same value as the function.

The function y = f (x) is a solution to the IVP

Approximate f (1.01).

We know f (1) = 2 and we want to estimate f (1.01). Sounds like Δ x = 0.01 and y_old = 2.

The differential equation tells us the derivative of y, so we can use the differential equation to figure out the slope of the tangent line. The slope of y at

(1, 2) is

So