Question

Find the linearization L(x) of f(x) at x = a

f(x)=√(4x+36), a=0

Answer 1

\[f(x) = \sqrt{4x+36}\]Can you find \(f'(x) =\)
?

Answer 2

its 1/(√x+9)

Answer 3

Then \[L(x) = f(a) + f'(a)(x - a)\]

Answer 4

so it would be f(0)+1/√(x+9)(x-a)?

Answer 5

Close:

\[L(x) = f(a) + f'(a)(x-a) = f(0) + f'(0)(x-0) = \]

Answer 6

You have to evaluate \(f'(0)\) to find the slope of the tangent line that linearizes the function at \(x = 0\). That isn't a variable, it's a fixed value.

Answer 7

This is essentially the \[y-y_0 = m(x-x_0)\]formula solved for \(y\) and cloaked in different clothing

Answer 8

ok

Answer 9

now what ?

Answer 10

well, what does that simplify to?

Answer 11

It's going to be the equation of a straight line...

Answer 12

at the point (0,6), the line will intersect the graph of \(f(x)\)

Answer 13

\[y = f(x) = \sqrt{4x+36}\]\[y_0 = f(x_0) = \sqrt{4x_0 + 36}\]\[x_0 = a\]\[y_0 = f(a) = \sqrt{4a+36} = \sqrt{4(0)+36} = 6\]
\[ f'(x) = \frac{1}{\sqrt{x+9}}\]\[m = f'(a) = \frac{1}{\sqrt{a+9}} = \frac{1}{\sqrt{0+9} }= \frac{1}{3}\]

So our line that is tangent to \(f(x) = \sqrt{4x+36}\) at \(x=a=0\) has slope \(m = \frac{1}{3}\) and passes through the point (0,6). That gives us

\[y - 6 = \frac{1}{3}(x-0)\] which I'm confident you can simplify and solve for \(y\)

Answer 14

ok thank you so much!

Answer 15

Looks like this:

Answer 16

You can see that for values of \(x\) close to \(a = 0\), the line and the curve are very close to each other. If you go away from \(a = 0\), the error (difference between the actual value of \(f(x)\) and the approximated value) increases. Here's another graph which shows the error as well (in olive):