Question

suppose that ( f ) is a function given as ( f(x) = \frac{1}{-x + 3} ). we will compute the derivative of ( f ) at ( x = 4 ) as follows. first, we compute and simplify the expression ( f(4 + h) ). ( f(4 + h) = ) then we compute and simplify the difference quotient, between ( x = 4 ) and ( x = 4 + h ). ( \frac{f(4 + h) - f(4)}{h} = ) the derivative of the function at ( x ) is the limit of the difference quotient as ( h ) approaches zero. ( f(4) = lim_{h \to 0} \frac{f(4 + h) - f(4)}{h} = )

Explanation:

Step1: Compute \( f(4 + h) \)

Substitute \( x = 4 + h \) into \( f(x)=\frac{1}{-x + 3} \).
\( f(4 + h)=\frac{1}{-(4 + h)+3}=\frac{1}{-4 - h + 3}=\frac{1}{-1 - h} \)

Step2: Compute \( f(4) \)

Substitute \( x = 4 \) into \( f(x)=\frac{1}{-x + 3} \).
\( f(4)=\frac{1}{-4 + 3}=\frac{1}{-1}=-1 \)

Step3: Compute the difference quotient \( \frac{f(4 + h)-f(4)}{h} \)

Substitute \( f(4 + h)=\frac{1}{-1 - h} \) and \( f(4)=-1 \) into the difference quotient.
\( \frac{\frac{1}{-1 - h}-(-1)}{h}=\frac{\frac{1}{-1 - h}+1}{h}=\frac{\frac{1 + (-1 - h)}{-1 - h}}{h}=\frac{\frac{1 - 1 - h}{-1 - h}}{h}=\frac{\frac{-h}{-1 - h}}{h}=\frac{-h}{h(-1 - h)}=\frac{1}{1 + h} \) (Note: \( h
eq0 \) for simplification)

Step4: Compute the limit as \( h\to0 \)

Find \( \lim_{h\to0}\frac{1}{1 + h} \). Substitute \( h = 0 \) into \( \frac{1}{1 + h} \).
\( \lim_{h\to0}\frac{1}{1 + h}=\frac{1}{1 + 0}=1 \)

Answer:

s:

• \( f(4 + h)=\boldsymbol{\frac{1}{-1 - h}} \)
• \( \frac{f(4 + h)-f(4)}{h}=\boldsymbol{\frac{1}{1 + h}} \) (for \( h

eq0 \))

• \( f'(4)=\boldsymbol{1} \)