How is this useful to us? Well, we can actually use this equation to approximate values of the function near point a. Take a look at this graph.
Notice that for x values near point a, we see that the function and the tangent line is relatively close to each other. Because of this, we are able to write that the function is approximately equal to the tangent line near point a. In other words,
where ≈ is the approximately symbol. This equation is known as the linear approximation formula. It is linear in a sense that the tangent is a straight line and we are using it to approximate the function. Using this approximation, we are able to approximate values that cannot be done by hand. For example, the square root of 2 or the natural log of 5 can all be approximated! One important thing to note is that this approximation only works for x values near point a. If you have a x value far from point a, then the approximation becomes really inaccurate.
Now don’t we take a look at a few examples of finding the linearization of a function and then look at how to use linear approximation!
Find the linearization of L(x) of the function at a
Question 1: Consider the function
Let’s say that we want to find the linearization of the function at point a=4.
To find the linearization L(x), recall that
Notice that in order to calculate L(x), we need f(a), f’(a) and a. Afterwards, were going to have to plug in everything in the formula to find L(x). Hence, I created these steps:
Step 1: Find a
Step 2: Find f(a)
Step 3: Find f’(a).
Step 4: Plug all three into the formula to find L(x)
Let’s follow these steps!
Step 1: Luckily a = 4 is given to us in the question, so we don’t have to look for it.
Know that the derivative of square roots is
And so plugging in x=a gives us:
Since we know a, f(a), and f’(a), we can now plug it into L(x) to find the linearization of f(x).
So L(x)=41x+1 is the linearization of this function at point x=4. In addition, it is also the tangent line of the function at point x=4.
How to do linear approximation
Remember earlier we said that we could use the equation of the tangent line to approximate values of the function near a? Let’s try this with the linearization we found earlier. Recall that
Now, let’s say I want to approximate f(4.04). If you were to plug this into the original function, then you would get √4.04 . This would be really hard to compute without a calculator. However, using linear approximation, we can say that
We just approximated the f(4.04) without a calculator! Now let’s actually see how close we were to the exact value of f(4.04). Notice that f(4.04) = √4.04 = 2.00997512422... So we are really close! We were only off when we got to the second decimal place!
Now so far, these questions gave us a function and a point to work with. What if none of these were given at all? What if the question only tells us to estimate a number?
Use Linear approximation to estimate a number
Suppose we want to estimate √10. How would we do it? We would need to use the linear approximation
Equation 3: Estimate with linear approx. pt.1
but we don’t even have a function and a point to work with. This means we have to make them ourselves. This leads us to do the following steps:
We know that linear approximation is just an estimation of the function’s value at a specified point. However, how do we know that if our estimation is an overestimate or an underestimate? We calculate the second derivative and look at the concavity.
Concave up vs Concave down
If the second derivative of the function is greater than 0 for values near a, then the function is concave up. This means that our approximation will be an underestimate. In other words,
Why? Let’s take a look at this graph.
Notice that f(x) is concave upward and the tangent line is right under f(x). Let’s say were to use the tangent line to approximate f(x). Then the y values of the tangent line are always going to be less than the actual value of f(x). Hence, we have an underestimate
Now if the second derivative of the function is less than 0 for values near a, then the function is concave down. This means that our approximation will be an overestimate. In other words,
Again, why? Let’s take a look at another graph.
Notice that f(x) is concave downward and the tangent line is right above f(x). Again, let’s say that we are going to use the tangent line to approximate f(x). Then the y values of the tangent line are always going to be greater than the actual value of f(x). Hence, we have an overestimate.
So if you ever need to see if your value is an underestimation or an overestimation, make sure you follow these steps:
Step 1: Find the second derivative
Step 2: look at the concavity of the function near point a
Step 3: Confirm that it is an underestimate/overestimate
Let’s take a look at an example:
Question 3: Let f(x) = √x and a = 4. If we linear approximate f(4.04), would it be an overestimate or an underestimate?
Step 1: See that
So the second derivative is
Notice that a=4, so we want to look at positive values of x near 4. Now look at the second derivative. When x is positive, we see that
Hence, it is concave down
We know that if the function is concave down, then the tangent line will be above the function. Hence, using the tangent line as an approximation will give an overestimated value.
Not only can we approximate values with linear approximation, but we can also approximate with differentials. To approximate, we use the following formula
where dy and dx are differentials, and f'(x) is the derivative of f in terms of x. Since we are dealing with very small changes in x and y, then we are going to use the fact that:
However, most of the questions we do involve setting
So using these facts will lead us to have:
This approximation is very useful when approximating the change of y. Keep in mind back then they didn’t have calculators, so this is the best approximation they could get for functions with square roots or natural logs.
Most of the time you will have to look for f’(x) and Δx yourself. In other words, follow these steps to approximate Δy!
Step 1: Find Δx
Step 2: Find f’(x)
Step 3: Plug everything into the formula to find dy. dy will be the approximation for Δy.
Let’s look at an example of using this approximation:
Question 4: Consider the function y = ln(x + 1). Suppose x changes from 0 to 0.01. Approximate Δy.
Step 1: Notice that x changes from 0 to 0.01, so the change in x would be:
The derivative would be:
Plugging everything in we have:
Hence, Δy ≈ 0.01
However, most of the time we want to estimate a value of the function, and not the change of the value. Hence we will add both sides of the equation by y, which gives us:
which is the same as:
This equation is a bit hard to read, so we are going to rearrange it even more. Let’s try to get rid of y and Δy. Notice that Δy+ y is basically the same as finding the value of the function at Δx+x. In other words,
Hence substituting this in our approximation above will give us:
where f(Δx+x) is value we are trying to estimate. How do we use this formula? I recommend following these steps:
Step 1: Set the number equal to f(Δx+x). Find Δx, x, and f(x).
Step 2: Calculate f’(x)
Step 3: Use the formula to approximate the number
Let’s use these steps for the following question.
Question 5: Use differentials to approximate √10.
Step 1: Compare f(Δx+x) with √10. Since √10 has a square root and 9 is a perfect square that is closest to 10, then let
See that there is no choice but to let Δx = 1
See that the derivative gives:
So this implies
Plugging everything into the formula gives us:
Hence, we just approximated the number.
One interesting thing to note is that linear approximation and differentials both give the same result for √10.
If you want to learn more about differentials, click this link:
Realize that the approximation becomes more and more accurate as we pick x values that are closer to a. In other words if we take the limit as x→a, then they will equal. So
Now we are going to put this aside and use it later, and actually look at l’hopital’s rule. We are going to assume a couple things here. Suppose that f(x) and g(x) are continuously differentiable at a real number a, f(a)=g(a)=0, and g’(a) ≠ 0. Then,
Now notice that we can apply the formula that we derived earlier right here. So now
Now instead of writing f’(a) and g’(a), we can apply limits as x→a (because we know f and g are differentiable). So
We always want to apply l’hoptial’s rule when we encounter indeterminate limits. There are two types of indeterminate forms. These indeterminate forms would be:
A lot of people make the mistake of using l’hopital’s rule without even checking if it is an indeterminate limit. So make sure you check it first! Otherwise, it will not work and you will get the wrong answer. Here is a guide to using l’hopital’s rule:
Step 1: Evaluate the limit directly.
Step 2: Check if it is one of the indeterminate forms. If it is, go to step 3.
Step 3: Use l’hopital’s rule.
Step 4: Check if you get another indeterminate form. Repeat Step 3 if you do.
Let’s take a look at a few examples using these steps.
Question 6: Evaluate the limit
Step 1: Evaluating the limit directly gives us
Yes, it is one of the indeterminate forms.
Applying l’hopital’s rule we have:
one is not an indeterminate form, so we are done and the answer is 1.
Now that question was a little bit easy, so why don’t we take a look at something that is a bit harder.
Question 7: Evaluate the limit
Step 1: Evaluating the limit directly we see that:
This is an indeterminate form, so go to step 3.
Applying l’hopital’s rule we have
This is another indeterminate form. So we have to go back to step 3 and apply l’hoptial’s rules again.
Applying l’hopital’s rule again we have:
Infinity is not an indeterminate form, so we are done and the answer is ∞
In this section, we will learn how to approximate unknown values of a function given known values using Linear Approximation. Linear Approximation has another name as Tangent Line Approximation because what we are really working with is the idea of local linearity, which means that if we zoom in really closely on a point along a curve, we will see a tiny line segment that has a slope equivalent to the slope of the tangent line at that point.
Intro: Linearization of f at a:L(x)=f(a)+f′(a)(x−a)
Consider the function f(x)=√x.
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