Question

Using the definition of the derivative, find f'(x). Then find f'(-5), f'(0) and f'(2) when the derivative exists.

$f(x)=\frac{75}{x}$

Please show all steps

Answer 1

Answer:

$f'(x)=-\dfrac{75}{x^2}$

f(-5) = -3, f(0) = Undefined, f(2) = -75/4

Step-by-step explanation:

To address the question of finding the derivative 'f'(x)' for the given function, we'll employ the definition of the derivative, which involves a limit process. This approach will not only help us determine the general formula for the derivative but also enable us to calculate specific values of the derivative at given points (f'(-5), f'(0) and f'(2)).

The definition of derivatives is given as:

$f'(x)=\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Our given function:

$f(x) = \dfrac{75}{x}$$\hrulefill$

Let's solve. Start by substituting f(x) into the definition:

$\Longrightarrow \lim_{h \to 0} \dfrac{\dfrac{75}{x+h}-\dfrac{75}{x}}{h}$

Find a common denominator to simplify the top of the fraction:

$\Longrightarrow \lim_{h \to 0} \dfrac{\dfrac{75}{x+h}\cdot\dfrac{x}{x}-\dfrac{75}{x}\cdot\dfrac{x+h}{x+h}}{h}$

$\Longrightarrow \lim_{h \to 0} \dfrac{\dfrac{75x}{x(x+h)}-\dfrac{75(x+h)}{x(x+h)}}{h}$

$\Longrightarrow \lim_{h \to 0} \dfrac{\dfrac{75x-75(x+h)}{x(x+h)}}{h}$

Continue simplifying:

$\Longrightarrow \lim_{h \to 0} \dfrac{75x-75(x+h)}{x(x+h)h}$

$\Longrightarrow \lim_{h \to 0} \dfrac{75x-75x-75h}{x^2h+xh^2}$

$\Longrightarrow \lim_{h \to 0} \dfrac{-75h}{h(x^2+xh)}\\\\\\\\\Longrightarrow \lim_{h \to 0} \dfrac{-75}{x^2+xh}$

Plug in the limit, h = 0:

$\Longrightarrow \lim_{h \to 0} \dfrac{-75}{x^2+x(0)}\\\\\\\\\therefore \boxed{f'(x)=-\dfrac{75}{x^2}}$

Thus, the derivative is found using the definition. $\hrulefill$

The calculated values for the derivative at the specified points are as follows:

When x = -5:

$\Longrightarrow f'(-5)=-\dfrac{75}{(-5)^2}\\\\\\\\\Longrightarrow f'(-5)=-\dfrac{75}{25}\\\\\\\\\therefore f'(-5) = \boxed{-3}$

When x = 0:

$\Longrightarrow f'(0)=-\dfrac{75}{(0)^2}\\\\\\\\\Longrightarrow f'(0)=-\dfrac{75}{0}\\\\\\\\\therefore f'(0) = \boxed{\text{Unde.}}$

When x = 2:

$\Longrightarrow f'(2)=-\dfrac{75}{(2)^2}\\\\\\\\\therefore \boxed{f'(2)=-\dfrac{75}{4}}$

Thus, all parts have been answered.