Question

question 5 (1 point)

use first principles to determine the instantaneous rate of change of \\( f(x) = \frac{1}{x} \\) at the point \\( \left(5, \frac{1}{5}\
ight) \\)

\\( \bigcirc \frac{1}{25} \\)

\\( \bigcirc -\frac{1}{25} \\)

\\( \bigcirc \frac{1}{5} \\)

\\( \bigcirc -\frac{1}{5} \\)

Explanation:

State the definition

To find the instantaneous rate of change of \(f(x) = \frac{1}{x}\) at \(x = 5\) using first principles, we use the Difference Quotient limit definition:

$$ f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h} $$

Substitute the function

Substitute \(f(x) = \frac{1}{x}\) into the limit expression:

$$ f'(5) = \lim_{h \to 0} \frac{\frac{1}{5+h} - \frac{1}{5}}{h} $$

Simplify the numerator

Find a common denominator for the fractions in the numerator:

$$ \frac{1}{5+h} - \frac{1}{5} = \frac{5 - (5+h)}{5(5+h)} = \frac{-h}{5(5+h)} $$

Divide by h

Substitute this simplified numerator back into the limit expression:

$$ f'(5) = \lim_{h \to 0} \frac{\frac{-h}{5(5+h)}}{h} = \lim_{h \to 0} \frac{-1}{5(5+h)} $$

Evaluate the limit

Evaluate the limit by direct substitution of \(h = 0\):

$$ f'(5) = \frac{-1}{5(5+0)} = -\frac{1}{25} $$

Answer:

• (A) \(\frac{1}{25}\)
• (B) \(-\frac{1}{25}\) (Correct answer)
• (C) \(\frac{1}{5}\)
• (D) \(-\frac{1}{5}\)