How do you find the tangent line approximation for #f(x)=sqrt(1+x)# near #x=0# ?

Question

Evelyn Agard · Accepted Answer

We need to find the derivative of #f(x)#.  We need to use the Chain Rule to find the derivative of #f(x)#. #f(x)=sqrt(1+x)=(1+x)^(1/2)# #f'(x)=(1/2)(1+x)^((1/2-1))*1# #f'(x)=(1/2)(1+x)^((1/2-2/2))# #f'(x)=(1/2)(1+x)^((-1/2))# #f'(x)=1/(2sqrt(1+x))# #f'(0)=1/(2sqrt(1+0))=1/(2sqrt(1))=1/2=0.5#

Adam Adkins · Answer

undefined To find the tangent line approximation for $ f(x) = \sqrt{1 + x} $ near $ x = 0 $, follow these steps:

1. Find the first derivative of $ f(x) $ using the power rule or chain rule.
$$ f'(x) = \frac{d}{dx}\sqrt{1+x} = \frac{1}{2\sqrt{1+x}} $$

2. Evaluate $ f'(x) $ at $ x = 0 $ to find the slope of the tangent line at that point.
$$ f'(0) = \frac{1}{2\sqrt{1+0}} = \frac{1}{2} $$

3. Use the point-slope form of a line equation, $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point of tangency and $ m $ is the slope of the tangent line.

4. Substitute $ x_1 = 0 $, $ y_1 = f(0) $, and $ m = f'(0) $ into the point-slope form.
$$ y - f(0) = f'(0)(x - 0) $$

5. Simplify the equation to find the tangent line approximation.
$$ y - 1 = \frac{1}{2}x $$

Samantha Adkison · Answer

undefined To find the tangent line approximation for $f(x) = \sqrt{1 + x}$ near $x = 0$, follow these steps:

1. Find the first derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$.
2. Evaluate $f'(0)$ to find the slope of the tangent line.
3. Use the point-slope form of a line to write the equation of the tangent line.

Here's how to do it:

1. Find the first derivative of $f(x)$:
$$f'(x) = \frac{d}{dx}\sqrt{1 + x}$$

2. Evaluate $f'(0)$:
$$f'(0) = \frac{d}{dx}\sqrt{1 + x}\bigg|_{x=0}$$

3. Use the point-slope form of a line to write the equation of the tangent line:
$$y - f(0) = f'(0)(x - 0)$$

4. Simplify the equation to obtain the tangent line approximation.

Following these steps:

1. The first derivative of $f(x) = \sqrt{1 + x}$ is:
$$f'(x) = \frac{1}{2\sqrt{1 + x}}$$

2. Evaluate $f'(0)$:
$$f'(0) = \frac{1}{2\sqrt{1 + 0}} = \frac{1}{2}$$

3. The point-slope form of the tangent line is:
$$y - f(0) = f'(0)(x - 0)$$
$$y - \sqrt{1 + 0} = \frac{1}{2}(x - 0)$$
$$y - 1 = \frac{1}{2}x$$

Therefore, the tangent line approximation for $f(x) = \sqrt{1 + x}$ near $x = 0$ is:
$$y = \frac{1}{2}x + 1$$