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# 6.4: Linear Approximations

Newton's method is one example of the usefulness of the tangent line as an approximation to a curve. Here we explore another such application. Recall that the tangent line to $$f(x)$$ at a point $$x=a$$ is given by $$L(x) = f'(a) (x-a) + f(a)$$. The tangent line in this context is also called the linear approximation to $$f$$ at $$a$$.

If $$f$$ is differentiable at $$a$$ then $$L$$ is a good approximation of $$f$$ so long as $$x$$ is "not too far'' from $$a$$. Put another way, if $$f$$ is differentiable at $$a$$ then under a microscope $$f$$ will look very much like a straight line. Figure $$\PageIndex{1}$$ shows a tangent line to $$y=x^2$$ at three different magnifications.

Figure $$\PageIndex{1}$$: The linear approximation to $$y=x^2$$

If we want to approximate $$f(b)$$, because computing it exactly is difficult, we can approximate the value using a linear approximation, provided that we can compute the tangent line at some $$a$$ close to $$b$$.

Example $$\PageIndex{1}$$

Let $$f(x)=\sqrt{x+4}$$. Then $$f'(x)=1/(2\sqrt{x+4})$$. The linear approximation to $$f$$ at $$x=5$$ is

$L(x)=1/(2\sqrt{5+4})(x-5)+\sqrt{5+4}=(x-5)/6+3.$

As an immediate application we can approximate square roots of numbers near 9 by hand. To estimate $$\sqrt{10}$$, we substitute 6 into the linear approximation instead of into $$f(x)$$, so

$\sqrt{6+4}\approx \dfrac{6-5}{6}+3 = \dfrac{19}{6} \approx 3.1 \overline{6}.$

This rounds to $$3.17$$ while the square root of 10 is actually $$3.16$$ to two decimal places, so this estimate is only accurate to one decimal place. This is not too surprising, as 10 is really not very close to 9; on the other hand, for many calculations, $$3.2$$ would be accurate enough.

### Contributors

• Integrated by Justin Marshall.