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Magazine Chapter 1: Problem 36
Let \(f(x)=\left\\{\begin{array}{rlr}x^{2} & \text { if } & x \leq-1 \\\
\sqrt{1-x^{2}} & \text { if } & -1
Short Answer
Expert verified
a) 4, b) 9, c) 0, d) 1, e) 1, f) 0.0447
Step by step solution
01
Determine the appropriate function for f(4)
Since 4 is greater than 1, we use the third piece of the piecewise function. For \(x > 1\), the function is \(f(x) = x\).
02
Calculate f(4)
Using the function \(f(x) = x\) for \(x > 1\), substituting 4 gives \(f(4) = 4\).
03
Determine the appropriate function for f(-3)
Since -3 is less than or equal to -1, we use the first piece of the piecewise function. For \(x \leq -1\), the function is \(f(x) = x^2\).
04
Calculate f(-3)
Using \(f(x) = x^2\) for \(x \leq -1\), substituting -3 gives \(f(-3) = (-3)^2 = 9\).
05
Determine the appropriate function for f(1)
Since 1 is between -1 and 1 (inclusive), we use the second piece of the piecewise function. For \(-1 < x \leq 1\), the function is \(f(x) = \sqrt{1-x^2}\).
06
Calculate f(1)
Using \(f(x) = \sqrt{1-x^2}\) for \(-1 < x \leq 1\), substituting 1 gives \(f(1) = \sqrt{1-1^2} = \sqrt{0} = 0\).
07
Determine the appropriate function for f(0)
Since 0 is between -1 and 1 (inclusive), we use the second piece of the piecewise function. For \(-1 < x \leq 1\), the function is \(f(x) = \sqrt{1-x^2}\).
08
Calculate f(0)
Using \(f(x) = \sqrt{1-x^2}\) for \(-1 < x \leq 1\), substituting 0 gives \(f(0) = \sqrt{1-0^2} = \sqrt{1} = 1\).
09
Determine the appropriate function for f(-1)
Since -1 is less than or equal to -1, we use the first piece of the piecewise function. For \(x \leq -1\), the function is \(f(x) = x^2\).
10
Calculate f(-1)
Using \(f(x) = x^2\) for \(x \leq -1\), substituting -1 gives \(f(-1) = (-1)^2 = 1\).
11
Determine the appropriate function for f(-0.999)
Since -0.999 is greater than -1 but less than or equal to 1, we use the second piece of the piecewise function. For \(-1 < x \leq 1\), the function is \(f(x) = \sqrt{1-x^2}\).
12
Calculate f(-0.999)
Using \(f(x) = \sqrt{1-x^2}\) for \(-1 < x \leq 1\), substituting -0.999 gives \(f(-0.999) = \sqrt{1-(-0.999)^2} \approx \sqrt{1-0.998001} \approx \sqrt{0.001999} \approx 0.0447\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Evaluation
Function evaluation involves finding the output of a function for a specific input. In our example, we have a piecewise function, which means it consists of multiple sub-functions, each applicable over a certain interval within the domain. To evaluate a specific input like in our exercise (for example, \(f(4)\)), you need to first determine which sub-function applies based on the input value. For \(f(4)\), since \(4 > 1\), we choose the function where \(f(x) = x\). Substituting \(x = 4\) into this rule yields \(f(4) = 4\). The key was identifying the correct piece of the piecewise function to use, which requires understanding the defined intervals of each sub-function.
Domain and Range
In mathematics, the domain refers to all possible input values of a function, while the range is all possible output values. For piecewise functions like the one in this exercise, the domain often has segments where different rules apply based on particular intervals.
- The domain for our example spans from negative to positive infinity, but different rules apply:
- For \(x \leq -1\), the function is \(f(x) = x^2\).
- For \(-1 < x \leq 1\), it is \(f(x) = \sqrt{1-x^2}\).
- For \(x > 1\), \(f(x) = x\).
Square Root Function
The square root function in the piecewise setup, \(f(x) = \sqrt{1-x^2}\), is defined for the interval \(-1 < x \leq 1\). This function is based on the Pythagorean identity of a circle with radius 1, centered at the origin. The output of this function gives the y-coordinate for a semi-circle's upper half.
- The square root function reaches its maximum value of 1 when \(x=0\).
- It drops to 0 as \(x\) approaches \(-1\) or \(1\).
Quadratic Function
Quadratic functions take the form \(f(x) = ax^2 + bx + c\), and are represented graphically as parabolas. In this exercise, the quadratic function is \(f(x) = x^2\) and applies when \(x \leq -1\). This part of the piecewise function opens upwards, since the coefficient of \(x^2\) is positive.
- As \(x\) becomes more negative, the value of \(f(x)\) increases.
- This results in a parabola that starts from "positive infinity" when moving leftward and goes on decreasing as it reaches x values closer to -1.
