Question

standard 1 homework
score: 23.5/28 answered: 18/20
question 19
determine and fully simplify the difference quotient, \\(\frac{f(a + h) - f(a)}{h}\\), for the function \\(f(x) = 3x^2 - 4x + 5\\).
\\(f(a+h) = \square\\)
\\(f(a) = \square\\)
\\(\frac{f(a+h) - f(a)}{h} = \square\\)
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Explanation:

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

Substitute \( x = a + h \) into \( f(x) = 3x^2 - 4x + 5 \).
\[

$$\begin{align*} f(a + h)&= 3(a + h)^2 - 4(a + h) + 5\\ &= 3(a^2 + 2ah + h^2) - 4a - 4h + 5\\ &= 3a^2 + 6ah + 3h^2 - 4a - 4h + 5 \end{align*}$$

\]

Step2: Find \( f(a) \)

Substitute \( x = a \) into \( f(x) = 3x^2 - 4x + 5 \).
\[
f(a)= 3a^2 - 4a + 5
\]

Step3: Find \( \frac{f(a + h) - f(a)}{h} \)

Subtract \( f(a) \) from \( f(a + h) \) and divide by \( h \).
\[

$$\begin{align*} \frac{f(a + h) - f(a)}{h}&=\frac{(3a^2 + 6ah + 3h^2 - 4a - 4h + 5)-(3a^2 - 4a + 5)}{h}\\ &=\frac{3a^2 + 6ah + 3h^2 - 4a - 4h + 5 - 3a^2 + 4a - 5}{h}\\ &=\frac{6ah + 3h^2 - 4h}{h}\\ &=\frac{h(6a + 3h - 4)}{h}\\ &= 6a + 3h - 4 \quad (h eq0) \end{align*}$$

\]

Answer:

\( f(a + h) = 3a^2 + 6ah + 3h^2 - 4a - 4h + 5 \)
\( f(a) = 3a^2 - 4a + 5 \)
\( \frac{f(a + h) - f(a)}{h} = 6a + 3h - 4 \)