Chapter 1: Problem 76
In Exercises \(53-76\), graph the piccewise-defined functions. State the domain and range in interval notation. Determine the intervals where the function is increasing, decreasing, or constant. $$f(x)=\left\\{\begin{array}{ll} x^{2} & x \leq-1 \\ x^{3} & -1< x <1 \\ x & x \geq 1 \end{array}\right.$$
Short Answer
Step by step solution
Label Each Piece of the Function
Determine the Domain
Determine the Range for Each Piece
Determine Intervals of Increase/Decrease/Constant
Sketch the Graph
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain and Range of Piecewise Functions
The range, however, is a bit more complex because it describes all the possible output values the function can produce. It's best to look at the range for each piece of the given function separately:
- The piece \(f(x) = x^2\) for \(x \leq -1\) means the function will output values starting from 1 (at \(x = -1\)) upwards indefinitely, forming the range \[\,[1, \infty)\,\].
- For \(f(x) = x^3\) with \(-1 < x < 1\), it spans output values between just below -1 and just below 1, creating the range \((-1, 1)\).
- The segment \(f(x) = x\) for \(x \geq 1\) results in outputs starting from 1 and going indefinitely upwards, thus, it's range is \[\,[1, \infty)\,\].
When merged together, the full range of the piecewise function is \([-1, \infty)\), considering all overlapping and individual ranges.
Increasing and Decreasing Intervals in Functions
Let's see how this applies:
- \(f(x) = x^2\) over \(x \leq -1\) is a decreasing function, as captured by its derivative, \(f'(x) = 2x\). Since \(x < 0\) in this interval, \(f'(x) < 0\), and the function decreases.
- \(f(x) = x^3\) over the interval \(-1 < x < 1\) is increasing. The derivative here, \(f'(x) = 3x^2\), is always positive, meaning that the function consistently grows in this piece.
- Lastly, \(f(x) = x\) for \(x \geq 1\) also shows an increasing pattern, due to its derivative, \(f'(x) = 1\), which is positive and constant.
Graphing Piecewise Functions
- For \(f(x) = x^2\) when \(x \leq -1\), plot the left part of a parabolic curve starting at \((-1, 1)\). This curve extends leftwards with a continuous decrease as \(x\) becomes more negative.
- For the piece \(f(x) = x^3\), sketch the segment going smoothly through the origin between \(-1 < x < 1\). This part of the graph resembles an 'S' shape pivoted at the origin due to the cubic nature.
- Finally, for \(f(x) = x\) over \(x \geq 1\), draw a straight line with a slope of 1 starting from the point \(1,1\) and moving diagonally upward.