How do you find the linearization at a=0 of #(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+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#

At #x=0#, we have #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)#