Back in the time of Newton and Leibniz, the calculus was unaided by calculators. Although mathematicians could quickly find derivatives using the various rules, the simple task of generating y-values from given x-values was incredibly difficult for the complexity of some functions. Thus, linearization was developed. This calculus tutorial will show you how linearization works, and how to apply it in a problem.

(In order for you to fully understand this tutorial, make sure you are comfortable deriving functions. Please check my article on this subject if you need help.)

What is linearization?

Before I show you the formula, let's see if we can get the concept down pat. Linearization works as an estimator for the value of a function at a certain x if we know one point along a function. Let us take, for example, the function y = 5x^{5} + 3x, where we know one point is the origin (0,0).

Now let's say we want to know the value of the function at 0.24. Back before the time of Texas Instruments, taking .24 to the fifth power would be quite a time-consuming task. What mathematicians instead proposed is that we use the tangent line at x = 0 to approximate the value of the function at x = 0.24. Of course, linearization only works for values relatively close to the point of tangency, since the gap between the line and the function will certainly increase greatly as x increases.

In this case, the actual value of the function at x = 0.24 is .7240, and linearization would approximate this to .72. Thus, mathematicians can relatively accurately find values of a function using calculus linearization to avoid laborious calculation. Let's see the math behind it.

The formula

Deriving the calculus formula for linearization shouldn't be too hard. All we're looking for is a tangent line to a point, and then plugging in an x-value near that point. If you need your memory refreshed on finding the equation of tangent lines, see my article on the subject. The linearization formula is just a slight reordering of the point-slope equation of the line:

where (a, f(a)) is the point of tangency of the approximating line, and f prime(a) is the slope of that line. Again, all you're doing is finding the tangent line of a point, and then using that line to approximate the value of the function of a nearby x-coordinate.

It should be noted that the function must be continuous on the closed interval and differentiable on the open interval for linearization to work at all points. However, if a function does not meet these criteria, linearization may still work for some points.

The problem

Let's do a problem involving linearization.

^{8}+ 44x

^{2}+ 100. Using linearization at x = 1, approximate the value of f at x = 1.02.

902 + 44 + 100 = 1046

Thus, the point of tangency is (1, 1046).

From here, we find f prime(a), or the slope of the tangent line, using the power rule for derivatives.

With these steps, we find that f prime(a) is 7304.

Now we have all we need to use the linearization formula to find the approximation at x = 1.02.

Thus, we conclude that the linear approximation, or linearization, of x = 1.02 is 1192.08. If you check for the real value of x = 1.02 in your calculator, you will find that this is off by a little over 10. Although this may seem like a large number, bear in mind that this function's rate of increase is astronomical, and a difference of 10 is therefore quite understandable.

Conclusion

And so comes the end of this calculus tutorial on using derivatives to find the linearization of of a function at a given point to approximate nearby points. This should prove useful as you go through your calculus course. Remember that a function must be continuous on the closed interval and differentiable on the open interval for this to always work. Good luck with your studies!

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