Chapter 1: Problem 4
Simplify each of the following difference quotients: a. \(\frac{(5+h)^{3}-125}{h}\) b. \(\frac{(3+h)^{4}-81}{h}\) c. \(\frac{\frac{1}{1+h}-1}{h}\) d. \(\frac{3(1+h)^{2}-3}{h}\) e. \(\frac{\frac{3}{4+h}-\frac{3}{4}}{h}\) f. \(\frac{\frac{-1}{2+h}+\frac{1}{2}}{h}\)
Short Answer
Expert verified
a. \(75 + 15h + h^2\), b. \(108 + 54h + 12h^2 + h^3\), c. \(-\frac{1}{1+h}\), d. \(6 + 3h\), e. \(-\frac{3}{16 + 4h}\), f. \(\frac{1}{4 + 2h}\).
Step by step solution
01
Simplify Part a
We start with the expression \( \frac{(5+h)^3 - 125}{h} \).First, expand \((5+h)^3\):\[(5+h)^3 = 5^3 + 3 \cdot 5^2 \cdot h + 3 \cdot 5 \cdot h^2 + h^3 = 125 + 75h + 15h^2 + h^3.\]Substituting back, the expression becomes:\[\frac{125 + 75h + 15h^2 + h^3 - 125}{h}.\]Simplify by cancelling out 125:\[\frac{75h + 15h^2 + h^3}{h}.\]Divide each term by \(h\):\[75 + 15h + h^2.\]Therefore, the simplified form is \(75 + 15h + h^2\).
02
Simplify Part b
Now for \( \frac{(3+h)^4 - 81}{h} \).Expand \((3+h)^4\):\[(3+h)^4 = 3^4 + 4 \cdot 3^3 \cdot h + 6 \cdot 3^2 \cdot h^2 + 4 \cdot 3 \cdot h^3 + h^4 = 81 + 108h + 54h^2 + 12h^3 + h^4.\]Substitute and simplify the expression:\[\frac{81 + 108h + 54h^2 + 12h^3 + h^4 - 81}{h} = \frac{108h + 54h^2 + 12h^3 + h^4}{h}.\]Divide each term by \(h\):\[108 + 54h + 12h^2 + h^3.\]Therefore, the simplified form is \(108 + 54h + 12h^2 + h^3\).
03
Simplify Part c
The expression to simplify is \( \frac{\frac{1}{1+h} - 1}{h} \).Rewrite \(1\) as \(\frac{1}{1+h}\):\[\frac{\frac{1}{1+h} - \frac{1+h}{1+h}}{h} = \frac{\frac{1 - (1+h)}{1+h}}{h}.\]This becomes:\[\frac{-h}{h(1+h)}.\]Cancel \(h\) in the numerator and denominator:\[\frac{-1}{1+h}.\]Therefore, the simplified form is \(\frac{-1}{1+h}\).
04
Simplify Part d
Let's simplify \( \frac{3(1+h)^2 - 3}{h} \).Expand \((1+h)^2\):\[(1+h)^2 = 1 + 2h + h^2.\]Substitute into the expression:\[\frac{3(1 + 2h + h^2) - 3}{h} = \frac{3 + 6h + 3h^2 - 3}{h}.\]Cancel the \(3\):\[\frac{6h + 3h^2}{h}.\]Divide each term by \(h\):\[6 + 3h.\]Thus, the simplified form is \(6 + 3h\).
05
Simplify Part e
Simplify \( \frac{\frac{3}{4+h} - \frac{3}{4}}{h} \).Find a common denominator for the fractions in the numerator:\[\frac{3(4) - 3(4+h)}{(4)(4+h)} = \frac{12 - (12 + 3h)}{16 + 4h}.\]Simplify the numerator:\[\frac{-3h}{16 + 4h}.\]Then the expression becomes:\[\frac{-3h}{h(16 + 4h)}.\]Cancel \(h\) in the numerator and denominator:\[-\frac{3}{16 + 4h}.\]Therefore, the simplified form is \(-\frac{3}{16 + 4h}\).
06
Simplify Part f
Finally, simplify \( \frac{\frac{-1}{2+h} + \frac{1}{2}}{h} \).Find a common denominator for the fractions in the numerator:\[\frac{-1(2) + 1(2+h)}{(2)(2+h)} = \frac{-2 + 2 + h}{4 + 2h}.\]Simplify the numerator:\[\frac{h}{4 + 2h}.\]So the expression becomes:\[\frac{h}{h(4 + 2h)}.\]Cancel \(h\) in the numerator and denominator:\[\frac{1}{4 + 2h}.\]Thus, the simplified form is \(\frac{1}{4 + 2h}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference Quotient
The difference quotient serves as a fundamental building block in calculus to understand how functions change. It provides the average rate of change of a function over a small interval, essentially giving us a way to approximate the slope of the secant line. For a function \( f(x) \), it is expressed as:
- \( \frac{f(x+h) - f(x)}{h} \)
Limits and Continuity
Understanding limits is essential since they allow us to make sense of expressions that might otherwise seem indeterminate, like those involving \( h \) approaching zero in a difference quotient. The limit is about exploring what happens to the function values as \( x \) gets arbitrarily close to a point, without necessarily reaching that point. For example, in the difference quotient \( \frac{f(x+h) - f(x)}{h} \), you consider the limit as \( h \) approaches zero:
- \( \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} = f'(x) \)
Algebraic Simplification
Simplifying algebraic expressions within difference quotients is a necessary skill to make limits and calculus calculations manageable. Preliminary steps often involve factoring, expanding polynomials, or finding common denominators to rewrite expressions in a way that cancels out the small variable \( h \), which might otherwise lead to undefined forms like \( \frac{0}{0} \). For instance:
- When expanding \((5+h)^3\), we use the binomial theorem: \(5^3 + 3 \cdot 5^2 \cdot h + 3 \cdot 5 \cdot h^2 + h^3\).
- After substituting terms back into the quotient, we can cancel terms to simplify like: \( \frac{75h + 15h^2 + h^3}{h} = 75 + 15h + h^2\).