Given the function f(x), the slope is evaluated by differentiating the f(x) function with respect to x.
Thus,
[tex]\text{slope = }\frac{df(x)}{dx}\text{ = f'(x)}[/tex]
Given a function f(x) as
[tex]f(x)\text{ = }\frac{4}{\sqrt[]{x+5}}[/tex]
Step 1:
Differentiate the f(x) function with respect to x.
[tex]\begin{gathered} f(x)\text{ = }\frac{4}{\sqrt[]{x+5}}\text{ can also be written as} \\ f(x)\text{ = }4(x+5)^{-\frac{1}{2}} \end{gathered}[/tex]
differentiating the f(x) function, we have
[tex]\begin{gathered} f(x)\text{ = }4(x+5)^{-\frac{1}{2}} \\ \frac{df(x)}{dx}\text{ = -}\frac{1}{2}\times4(x+5)^{-\frac{1}{2}-1} \\ \Rightarrow slope=f^{\prime}(x)=-2(x+5)^{-\frac{3}{2}} \end{gathered}[/tex]
Step 2:
Evaluate the slope at the point (-1, 1).
The slope at the point (-1, 1) is evaluated by substituting the values of x and y into the f'(x) function.
Thus,
[tex]\begin{gathered} x\text{ = -1, y = 1} \\ \text{substitute the values of x and y into f'(x)} \\ f^{\prime}(x)=-2(x+5)^{-\frac{3}{2}} \\ =-2(-1+5)^{-\frac{3}{2}} \\ =-2(4)^{-\frac{3}{2}} \\ =-2\times\frac{1}{8}^{} \\ =-\frac{1}{4} \end{gathered}[/tex]
Hence, the slope of the function at the point (-1, 1) is
[tex]-\frac{1}{4}[/tex]
