1.1 Q-2

Question Statement

Find the expression for $\frac{f(a+h) - f(a)}{h}$ and simplify where:

• (i) $f(x) = 6x - 9$
• (ii) $f(x) = \sin x$
• (iii) $f(x) = x^3 + 2x^2 - 1$
• (iv) $f(x) = \cos x$

Background and Explanation

To solve these types of problems, we need to apply the concept of the difference quotient, which is often used in the calculation of derivatives. The formula for the difference quotient is:

$$\frac{f(a+h) - f(a)}{h}$$

This expression represents the average rate of change of the function $f(x)$ over the interval from $a$ to $a+h$. To simplify it, we need to substitute the given function into this formula and simplify algebraically.

Solution

(i) For $f(x) = 6x - 9$:

1. Substitute $x = a + h$ into the function:

$$f(a+h) = 6(a+h) - 9 = 6a + 6h - 9$$

2. Now substitute $f(a+h)$ and $f(a)$ into the difference quotient formula:

$$\frac{f(a+h) - f(a)}{h} = \frac{6a + 6h - 9 - (6a - 9)}{h}$$

3. Simplify the expression:

$$= \frac{6a + 6h - 9 - 6a + 9}{h} = \frac{6h}{h}$$

4. Cancel out $h$:

$$= 6$$

So, the simplified result is 6.

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(ii) For $f(x) = \sin x$:

1. Substitute $x = a + h$ into the function:

$$f(a+h) = \sin(a + h)$$

2. Now substitute $f(a+h)$ and $f(a)$ into the difference quotient formula:

$$\frac{f(a+h) - f(a)}{h} = \frac{\sin(a + h) - \sin a}{h}$$

3. Use the identity for the difference of sines:

$$\sin \alpha - \sin \beta = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$$

Applying this identity:

$$= \frac{2}{h} \left[\cos \left(\frac{a+h+a}{2}\right) \sin \left(\frac{h}{2}\right)\right]$$

4. Simplify:

$$= \frac{2}{h} \cos \left(a + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$$

So, the simplified result is:

$$\frac{2}{h} \cos \left(a + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$$

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(iii) For $f(x) = x^3 + 2x^2 - 1$:

1. Substitute $x = a + h$ into the function:

$$f(a+h) = (a+h)^3 + (a+h)^2 - 1$$

2. Expand $(a+h)^3$ and $(a+h)^2$:

$$(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3$$ $$(a+h)^2 = a^2 + 2ah + h^2$$

3. Substitute these expansions into the difference quotient:

$$\frac{f(a+h) - f(a)}{h} = \frac{(a^3 + 3a^2h + 3ah^2 + h^3 + a^2 + 2ah + h^2 - 1) - (a^3 + 2a^2 - 1)}{h}$$

4. Simplify:

$$= \frac{a^3 + 3a^2h + 3ah^2 + h^3 + a^2 + 2ah + h^2 - 1 - a^3 - 2a^2 + 1}{h}$$

5. Combine like terms:

$$= \frac{h(3a^2 + 6ah + 3h^2 + 2a + h)}{h}$$

6. Cancel out $h$:

$$= 3a^2 + 6ah + 3h^2 + 2a + h$$

So, the simplified result is:

$$3a^2 + 6ah + 3h^2 + 2a + h$$

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(iv) For $f(x) = \cos x$:

1. Substitute $x = a + h$ into the function:

$$f(a+h) = \cos(a + h)$$

2. Now substitute $f(a+h)$ and $f(a)$ into the difference quotient formula:

$$\frac{f(a+h) - f(a)}{h} = \frac{\cos(a + h) - \cos a}{h}$$

3. Use the identity for the difference of cosines:

$$\cos \alpha - \cos \beta = -2 \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$$

Applying this identity:

$$= -\frac{2}{h} \left[\sin \left(\frac{a+h+a}{2}\right) \sin \left(\frac{h}{2}\right)\right]$$

4. Simplify:

$$= -\frac{2}{h} \sin \left(a + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$$

So, the simplified result is:

$$-\frac{2}{h} \sin \left(a + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$$

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Key Formulas or Methods Used

• Difference Quotient: $\frac{f(a+h) - f(a)}{h}$
• Sine and Cosine Difference Identities:
  • $\sin \alpha - \sin \beta = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$
  • $\cos \alpha - \cos \beta = -2 \sin \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$

Summary of Steps

1. Substitute $x = a + h$ into $f(x)$ to find $f(a+h)$.
2. Substitute $f(a+h)$ and $f(a)$ into the difference quotient formula.
3. Simplify the expression by expanding terms.
4. Apply trigonometric identities where needed (for sine and cosine).
5. Cancel out terms and simplify the expression to get the final result.