Question
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Consider the function \( f(x)=4 x^{2}-5 x-3 \) and complete parts (a) through (c). (a) Find \( f(a+h) \). (b) Find \( \frac{f(a+h)-f(a)}{h}, h \neq 0 \). (c) Find the instantaneous rate of change of \( f \) when \( a=7 \).

Ask by Norton Carrillo. in the United States
Jan 24,2025

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Answer

**(a)** \( f(a+h) = 4a^2 + 8ah + 4h^2 - 5a - 5h - 3 \) **(b)** \( \frac{f(a+h) - f(a)}{h} = 8a + 4h - 5 \) **(c)** The instantaneous rate of change at \( a = 7 \) is 51.

Solution

Let's analyze the function \( f(x) = 4x^2 - 5x - 3 \) and address each part of the problem step by step. ### Part (a): Find \( f(a+h) \) To find \( f(a+h) \), substitute \( x = a + h \) into the original function: \[ f(a+h) = 4(a + h)^2 - 5(a + h) - 3 \] Let's expand and simplify this expression: 1. **Expand \( (a + h)^2 \):** \[ (a + h)^2 = a^2 + 2ah + h^2 \] 2. **Multiply by 4:** \[ 4(a + h)^2 = 4a^2 + 8ah + 4h^2 \] 3. **Distribute the -5:** \[ -5(a + h) = -5a - 5h \] 4. **Combine all terms:** \[ f(a+h) = 4a^2 + 8ah + 4h^2 - 5a - 5h - 3 \] So, the expanded form of \( f(a+h) \) is: \[ f(a+h) = 4a^2 + 8ah + 4h^2 - 5a - 5h - 3 \] ### Part (b): Find \( \frac{f(a+h) - f(a)}{h} \), where \( h \neq 0 \) First, calculate \( f(a+h) - f(a) \): \[ f(a+h) - f(a) = (4a^2 + 8ah + 4h^2 - 5a - 5h - 3) - (4a^2 - 5a - 3) \] Simplify by subtracting \( f(a) \) from \( f(a+h) \): \[ f(a+h) - f(a) = 8ah + 4h^2 - 5h \] Now, divide by \( h \): \[ \frac{f(a+h) - f(a)}{h} = \frac{8ah + 4h^2 - 5h}{h} = 8a + 4h - 5 \] So, the difference quotient is: \[ \frac{f(a+h) - f(a)}{h} = 8a + 4h - 5 \] ### Part (c): Find the Instantaneous Rate of Change of \( f \) When \( a = 7 \) The instantaneous rate of change of \( f \) at \( a = 7 \) is the derivative of \( f \) at that point. From part (b), the derivative (as \( h \to 0 \)) is: \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = 8a - 5 \] Substitute \( a = 7 \): \[ f'(7) = 8(7) - 5 = 56 - 5 = 51 \] **Answer Summary:** **(a)** \( f(a+h) = 4a^{2} + 8a\,h + 4h^{2} - 5a - 5h - 3 \) **(b)** \( \frac{f(a+h)-f(a)}{h} = 8a + 4h - 5 \) **(c)** The instantaneous rate of change at \( a = 7 \) is 51.

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The Deep Dive

To get started with the function \( f(x) = 4x^2 - 5x - 3 \), let's break it down: (a) To find \( f(a+h) \), substitute \( a+h \) for \( x \) in the function: \[ f(a+h) = 4(a+h)^2 - 5(a+h) - 3 = 4(a^2 + 2ah + h^2) - 5a - 5h - 3 = 4a^2 + 8ah + 4h^2 - 5a - 5h - 3 \] (b) Now, compute \( \frac{f(a+h) - f(a)}{h} \): \[ f(a) = 4a^2 - 5a - 3 \] Thus, \[ f(a+h) - f(a) = (4a^2 + 8ah + 4h^2 - 5a - 5h - 3) - (4a^2 - 5a - 3) = 8ah + 4h^2 - 5h \] Therefore, \[ \frac{f(a+h) - f(a)}{h} = \frac{8ah + 4h^2 - 5h}{h} = 8a + 4h - 5 \] (c) For the instantaneous rate of change when \( a = 7 \), set \( a = 7 \) in the expression found in part (b): \[ \text{Instantaneous Rate} = 8(7) + 4h - 5 = 56 + 4h - 5 = 51 + 4h \] As \( h \) approaches \( 0 \), the instantaneous rate of change is \[ 51 \]

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