Linear approximation, sometimes called linearization, is one of the more useful applications of tangent line equations. We can use linear approximations to estimate the value of more complex functions. Coming up with a linear function that closely approximates another function at a certain point gives you something that is a lot easier to work with than the original function.

## How to find a linear approximation at a point

To find the linear approximation of a function at a point, you can simply use the formula for the linearization of a function. Let’s say you are given a function $$f(x)=x^3+2x$$ and you want to find its liner approximation at the point \(\mathbf{a=-2}\). You can do this by applying the formula

$$L(x)=f(a)+f'(a)(x-a).$$

The first step is to find the derivative of *f(x)*. In this case we can find *f'(x)* by using the power rule.

$$f'(x)=3x^2+2$$

Now that you know *f(x)*, *f'(x)*, and the value of *a*, you can plug all of this into the linearization formula above to find the linear approximation of *f(x)* near *x=a*.

$$L(x)=f(a)+f'(a)(x-a)$$ $$L(x)=\big( (-2)^3+2(-2) \big)+ \big( 3(-2)^2+2 \big) \big( x-(-2) \big)$$ $$L(x)=\big( -12 \big)+ \big( 14 \big) \big( x+2 \big)$$ $$L(x)=-12 + 14x+28$$ $$L(x)=14x+16$$

## Useful Applications of Linear Approximation

At this point you’re probably thinking, “this is just finding tangent line equations, who cares?”

But this does have some useful applications. One such application is using linear approximation to estimate the value of numbers that would be difficult or impossible to find without a calculator.

Let me show you what I mean by that.

### Example

Find the linearization of \(\mathbf{f(x)=\sqrt{x+1}}\) at a=0 and use it to approximate \(\mathbf{f(x)=\sqrt{0.9}}\).

Finding \(\mathbf{f(x)=\sqrt{0.9}}\) would be difficult without a calculator, right? But with linear approximation we can get pretty close by applying the process above and taking it a step further pretty easily.

We already have been given *f(x)* and the *a* value we need, so you just need to find *f'(x)* to apply our linearization formula.

$$f(x)=\sqrt{x+1}$$ $$f(x)=(x+1)^{\frac{1}{2}}$$

Writing *f(x)* in this form, we can simply apply the power rule and chain rule to find *f'(x)*.

$$f'(x)=\frac{1}{2}(x+1)^{-\frac{1}{2}}$$ $$f'(x)=\frac{1}{2\sqrt{x+1}}$$

Now that we know *f'(x)*, let’s first use that and *f(x)* to calculate *f(a)* and *f'(a)* before applying all of this to the linearization formula. Remember we were given *a=0*.

$$f(x) = \sqrt{x+1}$$ $$f(0)=\sqrt{0+1}$$ $$f(0)=\sqrt{1}$$ $$f(0)=1$$

And now you can find *f'(a)*, or *f'(0)*.

$$f'(x)=\frac{1}{2\sqrt{x+1}}$$ $$f'(0)=\frac{1}{2\sqrt{0+1}}$$ $$f'(0)=\frac{1}{2\sqrt{1}}$$ $$f'(0)=\frac{1}{2}$$

Now we can plug all of this into our linearization formula.

$$L(x)=f(0)+f'(0)(x-0)$$ $$L(x)=1+\frac{1}{2}x$$

#### How do you use this to estimate a number?

So we know that the function \(L(x)=1+\frac{1}{2}x\) is a good estimate of \(f(x)=\sqrt{x+1}\) when we plug in numbers close to *a=0* for *x*. And if we plug x=0 into f(x), that would give us \(f(0)=\sqrt{0+1}=\sqrt{1}\) which is close to the number we are trying to estimate, \(f(x)=\sqrt{0.9}\).

In reality, we could find the exact value of \(\sqrt{0.9}\) by evaluating \(f(-0.1)=\sqrt{-0.1+1}=\sqrt{0.9}\). But this is difficult, so instead we can use our linear approximation because we know it’s close to *f(x)* for *x* values near *x=0*. So instead of finding *f(-0.1)* we can find *L(-0.1)* and that will give us a good estimate.

$$L(-0.1)=1+\frac{1}{2}(-0.1)$$ $$L(-0.1)=1+(0.5)(-0.1)$$ $$L(-0.1)=1-0.05$$ $$L(-0.1)=0.95$$

So this tells us that \(\sqrt{0.9} \approx 0.95\).

## What if you aren’t given a function *f(x)* or an *x=a* value to use?

This seems nice to be able to estimate complicated values using this technique, but what if you need to estimate a complex value and aren’t given a function to use. Sometimes you aren’t given a function. And if you were trying to use this technique in the real world you probably wouldn’t be given a nice function to use and an *a* value that’s close to the input you need to estimate.

So how do you deal with this?

Well, the short answer is that you need to come up with the function and the *a* value on your own. Then once you have these you can apply the same technique above. Let me show you what I mean.

Let’s say you are told:

Use linear approximation to estimate \(\sqrt{24}\).

The first thing you want to do is come up with the function to use to apply the linearization formula to. Since we are trying to find \(\sqrt{24}\), our function is clearly going to need a square root in it somewhere. It is possible that you may need to experiment a bit and see how it works out, so let’s just start with the most obvious choice for now and say \(f(x)=\sqrt{x}\).

If we do say \(f(x)=\sqrt{x}\) then we can see that \(f(24)=\sqrt{24}\) which is exactly the number that we are trying to estimate. But f(24) is hard to evaluate. So to find the a value we will want to use, you should consider what number is near *24* that we could plug into the function \(f(x)=\sqrt{x}\) and easily figure out the result?

The closest number to 24 that we can easily plug into our function would be 25. Because 25 is a perfect square, we know that \(f(25)=\sqrt{25}=5\), which is a nice round number. This is perfect for linear approximation because we need an input that is near the location we are trying to estimate whose output we know.

Since *25 *is near *24 *and we know the exact value of *f(25)*, we can find the linear approximation of \(f(x)=\sqrt{x}\) at *a=25* and use this to estimate \(f(24)=\sqrt{24}\) just like we did in the example above.

I’m not going to work this all the way through because it will be the exact same process as the last example now that we know \(f(x)=\sqrt{x}\) and *a=25*. You can also see the rest of the solution to this problem in the video above.

## Differentials and how to find *dy*

Differentials go hand-in-hand with linear approximation and they have some interesting applications of their own.

Given some function *f(x)*, you can find its differential, *dy*, simply by using the following formula.

$$dy=f'(x) \ dx$$

For example, say you are given the function \(y=e^{\frac{x}{10}}\) and you are told to find the differential *dy*. All you need to do is apply the above formula.

You know that $$f(x)=e^{\frac{x}{10}}$$ Therefore, you can apply the chain rule to find $$f'(x)=e^{\frac{x}{10}} \cdot \frac{1}{10}$$ $$f'(x)=\frac{1}{10}e^{\frac{x}{10}}$$

So as a result, we know that

$$dy=\frac{1}{10}e^{\frac{x}{10}} \ dx$$

## Application of Differentials: Estimating Error

Estimating error is one of the more common uses for differentials. This can then be used to find the percentage error of a measurement. Let me explain what this means with an example.

Let’s say you measure a sphere to have a radius of *5 m*. However, you know that the instrument you used to measure the radius of this sphere could have up to *5 cm* (or *0.05 m*) of error. Use differentials to estimate the maximum error in the measured volume of this sphere.

Knowing the the volume of a sphere is $$V=\frac{4}{3}\pi r^3$$ we will use this as our function. We will be able to use the differential of this function to estimate the maximum possible error in the measured volume of this sphere based on the maximum possible error in the radius. We already know the possible error in the radius, so we can use this and relate it to the volume.

First all we need to do is find the differential of our volume equation. This will require taking the derivative of \(\mathbf{V=\frac{4}{3}\pi r^3}\). Using the formula of a differential as stated in the previous section, the differential is

$$dV=4 \pi r^2 \ dr$$

Then with this, all you need to do is plug in the information we know to find the possible error in the volume. Keep in mind, *r *represents the radius of the sphere that we know to be *5 m*. And *dr *will represent the possible error in the radius, which we know to be *0.05 m*. It’s important that we use the same units for all of these inputs since the output will use the same units in that case.

$$dV=4 \pi (5m)^2(0.05m)$$ $$dV=5 \pi m^3$$

So we know that the maximum error in the measurement we have for the volume of this sphere is \(\mathbf{5 \pi m^3\approx 15.708m^3}\).

## Percentage Error

Once we find the maximum possible error as outlined above, we can use this to find the percentage error of our measurement. This is simply the maximum possible error in the measurement divided by the measurement itself. So in this case it would be

$$Percentage \ Error = \frac{error \ in \ volume}{volume}.$$

$$Percentage \ Error = \frac{5 \pi}{\frac{500}{3}\pi} = 3\%$$