Linear Approximation Part IV

[Hckyplayer8]

Find the linearization of f(x)=tan x at a=0.

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

f(x)= tan x
f'(x)=sec²x

L(x)= tan 0 + sec²0 * ( x - 0)

Is everything setup properly? If so, what is the next step?

 

[Dr.Peterson]

What is tan(0)? What is sec(0)?

 

[Hckyplayer8]

What is tan(0)? What is sec(0)?

If a tangent line is one that touches a curve only once, and an angle of zero is just a line segment, then the tangent of zero is zero.

I had to look up the sec²0 equaling one. I'm a visual person and I was unable to visualize this portion of the problem.

 

[ksdhart2]

It might help to consider the definition of secant: $\sec(x) = \frac{1}{\cos(x)}$. If you can "visualize" that $\cos(0) = 1$, perhaps with the aid of the unit circle or some other method, then it directly follows that $\sec(0) = 1$ too.

As far as the problem goes, you can finish up by plugging in these values and simplifying a bit:

$\displaystyle L(x)= \tan(0) + \sec^2(0) * (x - 0) = 0 + 1(x - 0) = x$

In other words, for small values of $x$, we have $\tan(x) \approx x$

 

[Hckyplayer8]

Thank you both for posting.