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Chapter 1: Problem 57

Graph the function and determine the interval(s) for which \(f(x) \geq 0\). $$f(x)=9-x^{2}$$

Short Answer

Expert verified
The function \(f(x)=9-x^{2}\) is equal to or greater than zero for \(x\) in the interval \([-3 , 3]\).

Step by step solution

01

Understand the function and its graph

The function \(f(x)=9-x^{2}\) is a downward-opening parabola due to the negative sign before \(x^{2}\). The vertex of the parabola is at point (0, 9) since it's in the form \(f(x)=a-cx^{2}\), where a is the maximum value and c is a positive constant. The x-intercepts of this graph indicate when \(f(x)\) is zero.
02

Determine the x-intercepts

To find the x-intercepts, set \(f(x)\) equal to zero and solve for \(x\):\n\[9-x^{2} = 0\n\Rightarrow x^{2} = 9\n\Rightarrow x = \pm \sqrt{9}\n\Rightarrow x = \pm 3\] So, the x-intercepts of the function occur at \(x = -3\) and \(x = 3\). These points will be the boundaries for the interval for which \(f(x) \geq 0\).
03

Determine the intervals

Since the parabola opens downwards with its maximum at \(x=0\), the function will be zero or positive in the interval \([-3 , 3]\). Any value of \(x\) out of this interval will cause \(f(x)\) to be negative.

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Most popular questions from this chapter

For groups of 80 or more people, a charter bus company determines the rate per person according to the formula Rate \(=8-0.05(n-80), \quad n \geq 80\) where the rate is given in dollars and \(n\) is the number of people. (a) Write the revenue \(R\) for the bus company as a function of \(n\) (b) Use the function in part (a) to complete the table. What can you conclude? $$\begin{array}{|l|l|l|l|l|l|l|l|}\hline n & 90 & 100 & 110 & 120 & 130 & 140 & 150 \\\\\hline R(n) & & & & & & & \\\\\hline\end{array}$$

Use a graphing utility to graph the function. Be sure to choose an appropriate viewing window. $$f(x)=4+(1 / x)$$

Find the difference quotient and simplify your Answer: $$g(x)=\frac{1}{x^{2}}, \quad \frac{g(x)-g(3)}{x-3}, \quad x \neq 3$$

Consider \(f(x)=\sqrt{x-2}\) and \(g(x)=\sqrt[3]{x-2}\) Why are the domains of \(f\) and \(g\) different?

Sketch the graph of the function. $$g(x)=\left\\{\begin{array}{ll}x+6, & x \leq-4 \\\\\frac{1}{2} x-4, & x>-4\end{array}\right.$$
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