Answer :
To solve the given problem, we will proceed with finding the derivative of the function [tex]\( f(x) = \sqrt{x + 7} \)[/tex] and then determine the equation of the tangent line at the point [tex]\( a = 9 \)[/tex].
### Part a: Finding the Derivative
We start with the function [tex]\( f(x) = \sqrt{x + 7} \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we need to use the chain rule. Let's rewrite [tex]\( f(x) \)[/tex] using exponent notation:
[tex]\[ f(x) = (x + 7)^{1/2} \][/tex]
Using the chain rule for differentiation:
[tex]\[ \frac{d}{dx} \left[ (x + 7)^{1/2} \right] = \frac{1}{2} (x + 7)^{-1/2} \cdot \frac{d}{dx} (x + 7) \][/tex]
Since the derivative of [tex]\( x + 7 \)[/tex] is 1, we get:
[tex]\[ f'(x) = \frac{1}{2} (x + 7)^{-1/2} \][/tex]
Rewriting the expression back into its square root form, we have:
[tex]\[ f'(x) = \frac{1}{2\sqrt{x + 7}} \][/tex]
So, the derivative function [tex]\( f^{\prime}(x) \)[/tex] is:
[tex]\[ f^{\prime}(x) = \frac{1}{2\sqrt{x + 7}} \][/tex]
### Part b: Equation of the Tangent Line
Next, we need to find the equation of the tangent line to the function at [tex]\( a = 9 \)[/tex].
1. Find the slope of the tangent line at [tex]\( x = 9 \)[/tex]:
[tex]\[ f^{\prime}(9) = \frac{1}{2\sqrt{9 + 7}} = \frac{1}{2\sqrt{16}} = \frac{1}{2 \cdot 4} = \frac{1}{8} \][/tex]
2. Find the value of the function at [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = \sqrt{9 + 7} = \sqrt{16} = 4 \][/tex]
Now, we have a point on the graph [tex]\((9, 4)\)[/tex] and the slope of the tangent line [tex]\( \frac{1}{8} \)[/tex].
3. Using the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\((x_1, y_1) = (9, 4)\)[/tex] and [tex]\( m = \frac{1}{8} \)[/tex]:
[tex]\[ y - 4 = \frac{1}{8}(x - 9) \][/tex]
4. Simplifying the equation:
[tex]\[ y - 4 = \frac{1}{8}x - \frac{9}{8} \][/tex]
[tex]\[ y = \frac{1}{8}x - \frac{9}{8} + 4 \][/tex]
[tex]\[ y = \frac{1}{8}x - \frac{9}{8} + \frac{32}{8} \][/tex]
[tex]\[ y = \frac{1}{8}x + \frac{23}{8} \][/tex]
Thus, the equation of the tangent line at [tex]\( x = 9 \)[/tex] is:
[tex]\[ y = \frac{1}{8}x + \frac{23}{8} \][/tex]
To summarize,
- The derivative function [tex]\( f^{\prime}(x) \)[/tex] is: [tex]\[ \frac{1}{2\sqrt{x + 7}} \][/tex]
- The equation of the tangent line at [tex]\( x = 9 \)[/tex] is: [tex]\[ y = \frac{1}{8}x + \frac{23}{8} \][/tex]
### Part a: Finding the Derivative
We start with the function [tex]\( f(x) = \sqrt{x + 7} \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we need to use the chain rule. Let's rewrite [tex]\( f(x) \)[/tex] using exponent notation:
[tex]\[ f(x) = (x + 7)^{1/2} \][/tex]
Using the chain rule for differentiation:
[tex]\[ \frac{d}{dx} \left[ (x + 7)^{1/2} \right] = \frac{1}{2} (x + 7)^{-1/2} \cdot \frac{d}{dx} (x + 7) \][/tex]
Since the derivative of [tex]\( x + 7 \)[/tex] is 1, we get:
[tex]\[ f'(x) = \frac{1}{2} (x + 7)^{-1/2} \][/tex]
Rewriting the expression back into its square root form, we have:
[tex]\[ f'(x) = \frac{1}{2\sqrt{x + 7}} \][/tex]
So, the derivative function [tex]\( f^{\prime}(x) \)[/tex] is:
[tex]\[ f^{\prime}(x) = \frac{1}{2\sqrt{x + 7}} \][/tex]
### Part b: Equation of the Tangent Line
Next, we need to find the equation of the tangent line to the function at [tex]\( a = 9 \)[/tex].
1. Find the slope of the tangent line at [tex]\( x = 9 \)[/tex]:
[tex]\[ f^{\prime}(9) = \frac{1}{2\sqrt{9 + 7}} = \frac{1}{2\sqrt{16}} = \frac{1}{2 \cdot 4} = \frac{1}{8} \][/tex]
2. Find the value of the function at [tex]\( x = 9 \)[/tex]:
[tex]\[ f(9) = \sqrt{9 + 7} = \sqrt{16} = 4 \][/tex]
Now, we have a point on the graph [tex]\((9, 4)\)[/tex] and the slope of the tangent line [tex]\( \frac{1}{8} \)[/tex].
3. Using the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\((x_1, y_1) = (9, 4)\)[/tex] and [tex]\( m = \frac{1}{8} \)[/tex]:
[tex]\[ y - 4 = \frac{1}{8}(x - 9) \][/tex]
4. Simplifying the equation:
[tex]\[ y - 4 = \frac{1}{8}x - \frac{9}{8} \][/tex]
[tex]\[ y = \frac{1}{8}x - \frac{9}{8} + 4 \][/tex]
[tex]\[ y = \frac{1}{8}x - \frac{9}{8} + \frac{32}{8} \][/tex]
[tex]\[ y = \frac{1}{8}x + \frac{23}{8} \][/tex]
Thus, the equation of the tangent line at [tex]\( x = 9 \)[/tex] is:
[tex]\[ y = \frac{1}{8}x + \frac{23}{8} \][/tex]
To summarize,
- The derivative function [tex]\( f^{\prime}(x) \)[/tex] is: [tex]\[ \frac{1}{2\sqrt{x + 7}} \][/tex]
- The equation of the tangent line at [tex]\( x = 9 \)[/tex] is: [tex]\[ y = \frac{1}{8}x + \frac{23}{8} \][/tex]
