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Sagot :
Let's break down the solution into two parts: finding the derivative of [tex]\( f(x) \)[/tex] and then using this derivative to find the equation of the tangent line to the curve at the given point [tex]\((-5, -\frac{1}{9})\)[/tex].
### Part (a): Finding the Derivative [tex]\( f'(x) \)[/tex]
The given function is:
[tex]\[ f(x) = \frac{1}{x-4} \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = (x-4)^{-1} \][/tex]
Using the power rule of derivatives, we have:
[tex]\[ f'(x) = -1 \cdot (x-4)^{-2} \cdot \frac{d}{dx}(x-4) \][/tex]
Since [tex]\( \frac{d}{dx}(x-4) = 1 \)[/tex], the derivative simplifies to:
[tex]\[ f'(x) = -\frac{1}{(x-4)^2} \][/tex]
Thus:
[tex]\[ f'(x) = -\frac{1}{(x-4)^2} \][/tex]
### Part (b): Finding the Equation of the Tangent Line
We are given the point [tex]\((-5, -\frac{1}{9})\)[/tex].
First, calculate the slope of the tangent line at the given point using the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(-5) = -\frac{1}{(-5-4)^2} = -\frac{1}{81} = -0.012345679012345678 \][/tex]
The slope of the tangent line at [tex]\((-5, -\frac{1}{9})\)[/tex] is:
[tex]\[ \text{slope} = -0.012345679012345678 \][/tex]
Using the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the given point. Plugging in the values:
[tex]\[ y - \left(-\frac{1}{9}\right) = -0.012345679012345678(x - (-5)) \][/tex]
Simplify this equation step-by-step:
[tex]\[ y + \frac{1}{9} = -0.012345679012345678(x + 5) \][/tex]
We need [tex]\( \frac{1}{9} \approx 0.1111111111111111 \)[/tex]. So,
[tex]\[ y + 0.1111111111111111 = -0.012345679012345678(x + 5) \][/tex]
Next, distribute the slope:
[tex]\[ y + 0.1111111111111111 = -0.012345679012345678x - 0.06172839506172839 \][/tex]
[tex]\[ y = -0.012345679012345678x - 0.06172839506172839 - 0.1111111111111111 \][/tex]
Combining constants:
[tex]\[ y = -0.012345679012345678x - 0.1728395061728395 \][/tex]
Thus, the equation of the tangent line is:
[tex]\[ y = -0.012345679012345678x - 0.1728395061728395 \][/tex]
### Part (a): Finding the Derivative [tex]\( f'(x) \)[/tex]
The given function is:
[tex]\[ f(x) = \frac{1}{x-4} \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f(x) = (x-4)^{-1} \][/tex]
Using the power rule of derivatives, we have:
[tex]\[ f'(x) = -1 \cdot (x-4)^{-2} \cdot \frac{d}{dx}(x-4) \][/tex]
Since [tex]\( \frac{d}{dx}(x-4) = 1 \)[/tex], the derivative simplifies to:
[tex]\[ f'(x) = -\frac{1}{(x-4)^2} \][/tex]
Thus:
[tex]\[ f'(x) = -\frac{1}{(x-4)^2} \][/tex]
### Part (b): Finding the Equation of the Tangent Line
We are given the point [tex]\((-5, -\frac{1}{9})\)[/tex].
First, calculate the slope of the tangent line at the given point using the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(-5) = -\frac{1}{(-5-4)^2} = -\frac{1}{81} = -0.012345679012345678 \][/tex]
The slope of the tangent line at [tex]\((-5, -\frac{1}{9})\)[/tex] is:
[tex]\[ \text{slope} = -0.012345679012345678 \][/tex]
Using the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the given point. Plugging in the values:
[tex]\[ y - \left(-\frac{1}{9}\right) = -0.012345679012345678(x - (-5)) \][/tex]
Simplify this equation step-by-step:
[tex]\[ y + \frac{1}{9} = -0.012345679012345678(x + 5) \][/tex]
We need [tex]\( \frac{1}{9} \approx 0.1111111111111111 \)[/tex]. So,
[tex]\[ y + 0.1111111111111111 = -0.012345679012345678(x + 5) \][/tex]
Next, distribute the slope:
[tex]\[ y + 0.1111111111111111 = -0.012345679012345678x - 0.06172839506172839 \][/tex]
[tex]\[ y = -0.012345679012345678x - 0.06172839506172839 - 0.1111111111111111 \][/tex]
Combining constants:
[tex]\[ y = -0.012345679012345678x - 0.1728395061728395 \][/tex]
Thus, the equation of the tangent line is:
[tex]\[ y = -0.012345679012345678x - 0.1728395061728395 \][/tex]
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