Good Day,
For some reason, my brain is failing to craft a reasonable solution to this equation.
Question: Use the tangent line of $f(x) = \sqrt x$ at $a=16$ to approximate $\sqrt 17$
I believe we are looking to use linear approximation to discern the answer.
Initially I thought one could use the linear approximation formula to solve this:
$f'(a)(x-a) + f(a)$
This is where I started to get confused..
If we take the derivative of $\sqrt x$ we get: $1/2\sqrt x$
Following the formula that would give us:
$1/2\sqrt 16 = 1/8$ then $f(16) = 4$
Inserting the values into the formula: $1/8(x-16) + 4$
Now for x I was assuming we could use the $\sqrt 17$, however, that seems completely incorrect.
I am a little lost at what to perform next to gain an answer..
For example: $1/8(\sqrt 17 - 16) + 4 = 2.515$ which is certainly not the answer..
I apologize if I am totally missing the mark on this one.
I know that this does not have to do with formulating the tangent line equation and some linear approximation problems do not stump me in this fashion.. the square root values and only being provided a is tossing me about.
Any help is greatly appreciated!