Join IDNLearn.com and start getting the answers you've been searching for. Get the information you need from our experts, who provide reliable and detailed answers to all your questions.
Sagot :
To find the equation of the tangent line to the graph of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] at the point [tex]\((13, 2)\)[/tex], we need to follow several steps: find the derivative of [tex]\( f(x) \)[/tex], evaluate the derivative at [tex]\( x = 13 \)[/tex] to determine the slope, and then use the point-slope form of the equation of a line.
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
The function given is:
[tex]\[ f(x) = \sqrt{x - 9} \][/tex]
First, express the function as:
[tex]\[ f(x) = (x - 9)^{1/2} \][/tex]
To find the derivative of [tex]\( f(x) \)[/tex], use the power rule and the chain rule:
[tex]\[ f'(x) = \frac{1}{2}(x - 9)^{-1/2} \cdot (1) \][/tex]
[tex]\[ f'(x) = \frac{1}{2\sqrt{x - 9}} \][/tex]
### Step 2: Evaluate the derivative at [tex]\( x = 13 \)[/tex]
To find the slope of the tangent line at the point [tex]\((13, 2)\)[/tex], substitute [tex]\( x = 13 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(13) = \frac{1}{2\sqrt{13 - 9}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2\sqrt{4}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2 \cdot 2} \][/tex]
[tex]\[ f'(13) = \frac{1}{4} \][/tex]
Hence, the slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
### Step 3: Use the point-slope form of the equation of the line
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( (x_1, y_1) = (13, 2) \)[/tex] and the slope [tex]\( m = \frac{1}{4} \)[/tex]. Substituting these values, we get:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
### Step 4: Simplify the equation
To express the equation in the slope-intercept form [tex]\( y = mx + b \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
[tex]\[ y - 2 = \frac{1}{4}x - \frac{13}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + 2 \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + \frac{8}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
So, the equation of the tangent line to the graph of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] at the point [tex]\((13, 2)\)[/tex] is:
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
### Summary
- The derivative of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] is [tex]\( \frac{1}{2\sqrt{x - 9}} \)[/tex].
- The slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
- The equation of the tangent line at the point [tex]\((13, 2)\)[/tex] is [tex]\( y = \frac{1}{4}x - \frac{5}{4} \)[/tex].
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
The function given is:
[tex]\[ f(x) = \sqrt{x - 9} \][/tex]
First, express the function as:
[tex]\[ f(x) = (x - 9)^{1/2} \][/tex]
To find the derivative of [tex]\( f(x) \)[/tex], use the power rule and the chain rule:
[tex]\[ f'(x) = \frac{1}{2}(x - 9)^{-1/2} \cdot (1) \][/tex]
[tex]\[ f'(x) = \frac{1}{2\sqrt{x - 9}} \][/tex]
### Step 2: Evaluate the derivative at [tex]\( x = 13 \)[/tex]
To find the slope of the tangent line at the point [tex]\((13, 2)\)[/tex], substitute [tex]\( x = 13 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(13) = \frac{1}{2\sqrt{13 - 9}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2\sqrt{4}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2 \cdot 2} \][/tex]
[tex]\[ f'(13) = \frac{1}{4} \][/tex]
Hence, the slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
### Step 3: Use the point-slope form of the equation of the line
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( (x_1, y_1) = (13, 2) \)[/tex] and the slope [tex]\( m = \frac{1}{4} \)[/tex]. Substituting these values, we get:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
### Step 4: Simplify the equation
To express the equation in the slope-intercept form [tex]\( y = mx + b \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
[tex]\[ y - 2 = \frac{1}{4}x - \frac{13}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + 2 \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + \frac{8}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
So, the equation of the tangent line to the graph of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] at the point [tex]\((13, 2)\)[/tex] is:
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
### Summary
- The derivative of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] is [tex]\( \frac{1}{2\sqrt{x - 9}} \)[/tex].
- The slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
- The equation of the tangent line at the point [tex]\((13, 2)\)[/tex] is [tex]\( y = \frac{1}{4}x - \frac{5}{4} \)[/tex].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.
