2. Consider $$ f: x \rightarrow x^{2}-x $$ (a) Find the average rate of change of $$ f $$ over the interval $$ [1 ; 2] $$ ] (b) Find the instantaneous rate of ehange of $$ f $$ at $$ x=1 $$ by using first prineiples. (c) Determine $$ \lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h} $$ and interpret your answer. 3. Consider the function $$ f(x)=-2 a x^{2}+2 x $$ (a) Determine $$ \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $$ (b) Determine: (1) $$ f(3) $$ (3) the gradient of the tangent to $$ f $$ at $$ x=-4 $$ (4) the rate of change of $$ f $$ at $$ x=2 $$

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### Problem 2

#### (a) Find the average rate of change of $$ f $$ over the interval $$ [1, 2] $$

The average rate of change of a function $$ f $$ over the interval $$[a, b]$$ is given by the formula:

$$
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
$$

For $$ f(x) = x^2 - x $$, we need to calculate $$ f(1) $$ and $$ f(2) $$:

1. Calculate $$ f(1) $$:
   $$
   f(1) = 1^2 - 1 = 0
   $$

2. Calculate $$ f(2) $$:
   $$
   f(2) = 2^2 - 2 = 2
   $$

Now, substitute these values into the average rate of change formula:

$$
\text{Average Rate of Change} = \frac{f(2) - f(1)}{2 - 1} = \frac{2 - 0}{1} = 2
$$

#### (b) Find the instantaneous rate of change of $$ f $$ at $$ x=1 $$ using first principles

The instantaneous rate of change at a point $$ x $$ is given by the limit:

$$
\text{Instantaneous Rate of Change} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

For $$ x = 1 $$:

1. Calculate $$ f(1+h) $$:
   $$
   f(1+h) = (1+h)^2 - (1+h) = 1 + 2h + h^2 - 1 - h = h^2 + h
   $$

2. Now, substitute into the limit:
   $$
   \lim_{h \to 0} \frac{(h^2 + h) - 0}{h} = \lim_{h \to 0} \frac{h^2 + h}{h} = \lim_{h \to 0} (h + 1) = 1
   $$

#### (c) Determine $$ \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} $$

1. Calculate $$ f(3) $$:
   $$
   f(3) = 3^2 - 3 = 6
   $$

2. Calculate $$ f(3+h) $$:
   $$
   f(3+h) = (3+h)^2 - (3+h) = 9 + 6h + h^2 - 3 - h = h^2 + 5h + 6
   $$

3. Substitute into the limit:
   $$
   \lim_{h \to 0} \frac{(h^2 + 5h + 6) - 6}{h} = \lim_{h \to 0} \frac{h^2 + 5h}{h} = \lim_{h \to 0} (h + 5) = 5
   $$

### Problem 3

#### (a) Determine $$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

For $$ f(x) = -2ax^2 + 2x $$:

1. Calculate $$ f(x+h) $$:
   $$
   f(x+h) = -2a(x+h)^2 + 2(x+h) = -2a(x^2 + 2xh + h^2) + 2x + 2h = -2ax^2 - 4axh - 2ah^2 + 2x + 2h
   $$

2. Substitute into the limit:
   $$
   \lim_{h \to 0} \frac{(-2ax^2 - 4axh - 2ah^2 + 2x + 2h) - (-2ax^2 + 2x)}{h} = \lim_{h \to 0} \frac{-4axh - 2ah^2 + 2h}{h}
   $$
   $$
   = \lim_{h \to 0} (-4ax - 2ah + 2) = -4ax + 2
   $$

#### (b) Determine:
1. $$ f(3) $$:
   $$
   f(3) = -2a(3^2) + 2(3) = -18a + 6
   $$

2. The gradient of the tangent to $$ f $$ at $$ x=-4 $$:
   $$
   \text{Gradient} = -4a(-4) + 2 = 16a + 2
   $$

3. The rate of change of $$ f $$ at $$ x=2 $$:
   $$
   \text{Rate of Change} = -4a(2) + 2 = -8a + 2
   $$

### Summary of Results
- Average rate of change over $$[1, 2]$$: $$ 2 $$
- Instantaneous rate of change at $$ x=1 $$: $$ 1 $$
- Limit for $$ x=3 $$: $$ 5 $$
- $$ f(3) $$: $$ -18a + 6 $$
- Gradient at $$ x=-4 $$: $$ 16a + 2 $$
- Rate of change at $$ x=2 $$: $$ -8a + 2 $$
- **Problem 2:** - (a) Average rate of change over $$[1, 2]$$: 2 - (b) Instantaneous rate of change at $$ x=1 $$: 1 - (c) Limit at $$ x=3 $$: 5 - **Problem 3:** - (a) Limit as $$ h \to 0 $$ of $$ \frac{f(x+h) - f(x)}{h} $$: $$ -4ax + 2 $$ - (b) - $$ f(3) $$: $$ -18a + 6 $$ - Gradient at $$ x=-4 $$: $$ 16a + 2 $$ - Rate of change at $$ x=2 $$: $$ -8a + 2 $$

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Problem 2

(a) Find the average rate of change of over the interval

(b) Find the instantaneous rate of change of at using first principles

© Determine

Problem 3

(a) Determine

(b) Determine:

Summary of Results

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