Question

2.3 part ii homework - derivatives using limits
score: 20/80 answered: 2/8
question 3
use the four - step definition of the derivative to find $f(x)$ if $f(x)=3x - 14$.
$f(x + h)=\square$
$f(x + h)-f(x)=\square$
$\frac{f(x + h)-f(x)}{h}=\square$
find $f(x)$ by determining $\lim\limits_{h\to 0}\frac{f(x + h)-f(x)}{h}=\square$
question help: video

Explanation:

Step1: Find \( f(x + h) \)

Substitute \( x + h \) into \( f(x) = 3x - 14 \). So \( f(x + h)=3(x + h)-14 = 3x+3h - 14 \).

Step2: Calculate \( f(x + h)-f(x) \)

Substitute \( f(x + h)=3x + 3h-14 \) and \( f(x)=3x - 14 \) into the expression. Then \( f(x + h)-f(x)=(3x + 3h-14)-(3x - 14)=3x+3h - 14 - 3x + 14 = 3h \).

Step3: Simplify \( \frac{f(x + h)-f(x)}{h} \)

Substitute \( f(x + h)-f(x)=3h \) into the fraction. So \( \frac{f(x + h)-f(x)}{h}=\frac{3h}{h}=3 \) (where \( h
eq0 \)).

Step4: Find the limit as \( h

ightarrow0 \)
We need to find \( \lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h} \). Since \( \frac{f(x + h)-f(x)}{h}=3 \) (for \( h
eq0 \)), the limit as \( h
ightarrow0 \) is also 3.

Answer:

s (for each blank in order):
\( f(x + h)=\boldsymbol{3x + 3h - 14} \)

\( f(x + h)-f(x)=\boldsymbol{3h} \)

\( \frac{f(x + h)-f(x)}{h}=\boldsymbol{3} \)

\( f'(x)=\boldsymbol{3} \)