How do you find the linearization at a=0 of ${(2 + 8 x)}^{\frac{1}{2}}$ $(2+8x)^(1/2)$ ?

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Answer 1 · 2015-11-06T20:48:05

The linearization is an equation for the tangent line.

$f (x) = {(2 + 8 x)}^{\frac{1}{2}} = \sqrt{2 + 8 x}$ $f(x) = (2+8x)^(1/2) = sqrt(2+8x)$

Find an equation for the line tangent to the graph of the function at the point where $x = a = 0$ $x=a=0$

At $x = 0$ $x=0$ , we have $y = f (0) = \sqrt{2}$ $y = f(0) = sqrt2$

$f'(x) = 4(2+8x)^(-1/2) = 4/sqrt(2+8x)$

At the point $(0,sqrt2)$ , the slope of the tangent is $m = f'(2) = 2sqrt2$

So the tangent line has point slope equation:

$y-sqrt2 = 2sqrt2(x-0)$ $" "$ $" "$ $y-f(a)=f'(a)(x-a)$

And the linearization is

$L(x) = sqrt2 + 2sqrt2x$ $" "$ $" "$ $L(x) = f(a)+f'(a)(x-a)$

How do you find the linearization at a=0 of (2+8x)^(1/2)(2+8x)12(2+8x)^(1/2)?