Question

part b: written response show all work for full marks.

1. if $f(x)=3x^{2}-2x$, then use first principles to show that $f(x)=6x-2$.

$y=\lim_{h \to 0} \frac{f(a+h) - f(x)}{h}$
12 $y=\lim_{h \to 0} \frac{3(a+h)^{2}-2(a+h)-3x^{2}-2x}{h}$

Explanation:

Step1: Recall first principles formula

The derivative of a function $f(x)$ from first principles is given by:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step2: Substitute $f(x+h)$ and $f(x)$

Given $f(x)=3x^2-2x$, calculate $f(x+h)=3(x+h)^2-2(x+h)$. Substitute into the formula:
$$f'(x) = \lim_{h \to 0} \frac{3(x+h)^2 - 2(x+h) - (3x^2 - 2x)}{h}$$

Step3: Expand and simplify numerator

Expand $(x+h)^2 = x^2 + 2xh + h^2$, then distribute and combine like terms:

$$\begin{align*} \text{Numerator} &= 3(x^2+2xh+h^2) -2x -2h -3x^2 +2x\\ &= 3x^2 +6xh +3h^2 -2x -2h -3x^2 +2x\\ &= 6xh + 3h^2 -2h \end{align*}$$

Now the expression becomes:
$$f'(x) = \lim_{h \to 0} \frac{6xh + 3h^2 -2h}{h}$$

Step4: Factor and cancel $h$

Factor $h$ from the numerator, then cancel $h$ (since $h
eq 0$ as $h \to 0$):
$$f'(x) = \lim_{h \to 0} \frac{h(6x + 3h -2)}{h} = \lim_{h \to 0} (6x + 3h -2)$$

Step5: Evaluate the limit

Substitute $h=0$ into the simplified expression:
$$f'(x) = 6x + 3(0) -2 = 6x -2$$

Answer:

$f'(x)=6x-2$