3.10: Linear Approximations

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Consider $f(x) = \sqrt [3]{x}$ . The equation of the tangent line to $y = f(x)$ at the point $(8,f(8))$ is: $y = \answer {\frac {1}{12}x+\frac {4}{3}}.$

Use the equation of the tangent line from the previous part to approximate $\sqrt [3]{7.99}$ .

The equation from the previous part was $y = \frac {1}{12}(x-8)+2.$ Therefore, if we plug in $x=7.99$ , we will get an approximation for $f(7.99) = \sqrt [3]{7.99}$ . Plugging in $x=7.99$ into the tangent line equation gives: $f(7.99) \approx \answer {-\frac {1}{1200}+2}.$

Consider the quantity $(1.01)^{10}$ .

We can approximate this quantity by using the tangent line to $y = x^{10}$ at the point $x=1$ . The equation of the line is $y = \answer {10x-9}.$
Using the tangent line from the previous part, we get an approximation of $\answer {1.1}$ .
By examining the graph of $y =x^{10}$ and the tangent line to $(1,1)$ , is this answer an over-or under-approximation?
over-approximation under-approximation
Just to note, the actual value is (to five decimal places): $1.10462$ .

Suppose the graph of $y= f(x)$ has a horizontal tangent line at the point $(1,3)$ .

Which of the following is the equation of the tangent line to $y=f(x)$ at $(1,3)$ ?
$y=x+2$ $y=2x+1$ $y=3$ $y=-x+4$
Using a linear approximation, we can say $f(.99) \approx \answer {3}$ .

Suppose $f(x)$ satisfies $f(2)= -3$ and $f'(x) = \sqrt {12+x^2}$ . Using a linear approximation, we can say $f(2.1) \approx \answer {-2.6}.$

A linear approximation for $f(x) = \tan (x)$ tells us that $\tan \left (\frac {\pi }{4}+.01 \right ) \approx \answer {1.02}.$

Is this an over-estimate or under-estimate?

over-approximation under-approximation