Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Let [tex]\( f(x) = \frac{1}{x-4} \)[/tex].

(a) Find the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = -\frac{1}{(x-4)^2} \][/tex]

(b) Find an equation of the tangent line to the curve at the point [tex]\((-5, -\frac{1}{9})\)[/tex]:
[tex]\[ y = \square \][/tex]


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]