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

Question

Charles Arocha · Accepted Answer

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)#

Cameron Alexander · Answer

undefined To find the linearization at $ a = 0 $ of $ (2 + 8x)^{\frac{1}{2}} $, we need to find the first-order Taylor polynomial centered at $ a = 0 $.

First, find the derivative of $ (2 + 8x)^{\frac{1}{2}} $ with respect to $ x $:

$$ \frac{d}{dx}\left((2 + 8x)^{\frac{1}{2}}\right) = \frac{1}{2}(2 + 8x)^{-\frac{1}{2}}(8) = \frac{4}{\sqrt{2 + 8x}} $$

Now, evaluate the function and its derivative at $ x = 0 $:

$$ f(0) = (2 + 8(0))^{\frac{1}{2}} = 2^{\frac{1}{2}} = \sqrt{2} $$

$$ f'(0) = \frac{4}{\sqrt{2 + 8(0)}} = \frac{4}{\sqrt{2}} $$

The linearization of $ (2 + 8x)^{\frac{1}{2}} $ at $ a = 0 $ is given by:

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

Substitute $ f(0) = \sqrt{2} $, $ f'(0) = \frac{4}{\sqrt{2}} $, and $ a = 0 $ into the equation:

$$ L(x) = \sqrt{2} + \frac{4}{\sqrt{2}}(x - 0) = \sqrt{2} + 2\sqrt{2}x = (1 + 2x)\sqrt{2} $$

Therefore, the linearization at $ a = 0 $ of $ (2 + 8x)^{\frac{1}{2}} $ is $ L(x) = (1 + 2x)\sqrt{2} $.