Question

Calculate \(f(x+h)\) when (a) \(f(x)=x^{2}\) (b) \(f(x)=x^{3}\) (c) \(f(x)=\frac{1}{x}\) In each case write down the corresponding expression for \(f(x+h)-f(x)\).

Short answer

Question: Calculate \(f(x+h)\) and the expression for \(f(x+h)-f(x)\) for the following functions: a) \(f(x) = x^2\) b) \(f(x) = x^3\) c) \(f(x) = \frac{1}{x}\) Answer: a) For \(f(x) = x^2\), \(f(x+h) = x^2 + 2xh + h^2\) and \(f(x+h) - f(x) = 2xh + h^2\). b) For \(f(x) = x^3\), \(f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3\) and \(f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3\). c) For \(f(x) = \frac{1}{x}\), \(f(x+h) = \frac{1}{x+h}\) and \(f(x+h) - f(x) = \frac{-h}{x(x+h)}\).

Step-by-step solution

(Step 1: Calculate f(x+h) for f(x) = x^2)

Substitute \(x\) with \(x+h\) in the function \(f(x)=x^2\). Then, \(f(x+h)=(x+h)^2\) Now expand the expression: \(f(x+h) = x^2 + 2xh + h^2\)

(Step 2: Calculate f(x+h) - f(x) for f(x) = x^2)

Subtract \(f(x)\) from \(f(x+h)\): \(f(x+h)-f(x) = (x^2 + 2xh + h^2) - x^2\) Simplify the expression: \(f(x+h)-f(x) = 2xh + h^2\) For the function \(f(x)=x^2\), \(f(x+h)=x^2+2xh+h^2\) and \(f(x+h)-f(x) = 2xh + h^2\).

(Step 3: Calculate f(x+h) for f(x) = x^3)

Substitute \(x\) with \(x+h\) in the function \(f(x)=x^3\). Then, \(f(x+h)=(x+h)^3\) Expand the expression: \(f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3\)

(Step 4: Calculate f(x+h) - f(x) for f(x) = x^3)

Subtract \(f(x)\) from \(f(x+h)\): \(f(x+h)-f(x) = (x^3 + 3x^2h + 3xh^2 + h^3) - x^3\) Simplify the expression: \(f(x+h)-f(x) = 3x^2h + 3xh^2 + h^3\) For the function \(f(x)=x^3\), \(f(x+h)=x^3+3x^2h+3xh^2+h^3\) and \(f(x+h)-f(x) = 3x^2h + 3xh^2 + h^3\).

(Step 5: Calculate f(x+h) for f(x) = 1/x)

Substitute \(x\) with \(x+h\) in the function \(f(x)=\frac{1}{x}\). Then, \(f(x+h)=\frac{1}{x+h}\)

(Step 6: Calculate f(x+h) - f(x) for f(x) = 1/x)

Subtract \(f(x)\) from \(f(x+h)\): \(f(x+h)-f(x) = \frac{1}{x+h} - \frac{1}{x}\) Find a common denominator to combine the fractions: \(f(x+h)-f(x) = \frac{x-(x+h)}{x(x+h)}\) Simplify the expression: \(f(x+h)-f(x) = \frac{-h}{x(x+h)}\) For the function \(f(x)=\frac{1}{x}\), \(f(x+h)=\frac{1}{x+h}\) and \(f(x+h)-f(x) = \frac{-h}{x(x+h)}\).

Key concepts

Functions

In mathematics, a function is basically a rule that relates each input to exactly one output. When you see a function notation like \(f(x)\), it signifies that for every value of \(x\), there is a specific result. Functions are essential in calculus because they allow us to explore how quantities change in relation to each other. They can be expressed in various forms like linear \(f(x) = x + 2\), quadratic \(f(x) = x^2\), cubic \(f(x) = x^3\), or even rational functions \(f(x) = \frac{1}{x}\).
Understanding functions and their properties is foundational because many calculus operations are based on how these functions behave.

Limits

The concept of limits helps mathematicians understand the behavior of functions as inputs approach a certain value. Limits examine what happens to \(f(x)\) as \(x\) approaches a point, which is particularly useful when \(x\) is approaching a value where the function is not clearly defined.
For example, when calculating limits, expressions like \(f(x+h) - f(x)\) come into play, showing us how small changes in \(x\) affect the function's output. This is crucial in calculus, as it lays the groundwork for defining derivatives.

Algebra

Algebra involves working with symbols and numbers to solve equations or simplify expressions. It is a cornerstone of calculus since many problems require manipulation of algebraic expressions.
For instance, in the exercises given, algebra is used to expand expressions like \((x+h)^2\) or \((x+h)^3\), showing the power and utility of algebra in calculus. This is particularly useful when finding \(f(x+h)\) and calculating differences such as \(f(x+h)-f(x)\) to simplify results or to set up for further calculus computations.

Differentiation

Differentiation is one of the fundamental concepts in calculus. It involves finding the derivative of a function, which tells us how the function changes as its input changes. Derivatives are particularly useful in determining things like slope, velocity, or any situation involving rates of change.
In our exercises, calculating \(f(x+h) - f(x)\) prepares us for differentiation by examining the incremental changes in function values. This step is crucial as differentiation relies on finding the limit of these differences as \(h\) approaches zero. Ultimately, understanding how to derive these changes lets us solve practical problems involving real-world scenarios of change.