How do you find the linearization at a=16 of #f(x) = x^(1/2)#?

Question

Samantha Adkison · Accepted Answer

Use the formula #L(x)=f(a)+f'(a)(x-a)# to get #L(x)=4+1/8(x-16)=1/8x+2# as the linearization of #f(x)=x^{1/2}# at #a=16#.

For #f(x)=x^{1/2}# we have #f'(x)=1/2 x^{-1/2}# so that #f(a)=f(16)=16^{1/2}=4# and #f'(a)=f'(16)=1/2 * 16^{-1/2}=1/2 * 1/4 = 1/8#. Therefore, the function #L(x)=f(a)+f'(a)(x-a)=4+1/8(x-16)=1/8x+2# is the linearization of #f(x)=x^{1/2}# at #a=16#. This can be used to obtain good approximations to square roots of numbers near 16. For example, #sqrt{16.5}=f(16.5) approx L(16.4)=4+1/8(16.5-16)=4+1/8 * 1/2 = 4+ 1/16=4.0625#. A more accurate approximation with technology is: #sqrt{16.5} approx 4.062019202#. The error in our approximation is about #4.062019202-4.0625 approx -0.00048#, which is pretty good.

Aurora Alvarado · Answer

undefined To find the linearization of $f(x) = \sqrt{x}$ at $a = 16$, we follow these steps:

1. Find the first derivative of $f(x)$ with respect to $x$.
$$f'(x) = \frac{1}{2\sqrt{x}}$$

2. Evaluate $f'(x)$ at $x = a$.
$$f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8}$$

3. Write the equation of the tangent line using the point-slope form:
$$y - f(a) = f'(a)(x - a)$$

4. Substitute the values of $a$, $f(a)$, and $f'(a)$.
$$y - f(16) = \frac{1}{8}(x - 16)$$

5. Simplify the equation to obtain the linearization.
$$y - 4 = \frac{1}{8}(x - 16)$$