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Chapter 1: Problem 125
Prove or disprove: If \(x\) and \(y\) are real numbers with \(y \geq 0\) and \(y(y+1) \leq(x+1)^{2},\) then \(y(y-1) \leq x^{2}\)
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Testing for Continuity In Exercises \(77-84\) , describe the interval(s) on which the function is continuous. $$ f(x)=\cos \frac{1}{x} $$
Finding a Limit In Exercises \(7-26\) , find the limit (if it exists). If it does not exist, explain why. $$ \lim _{x \rightarrow 1} f(x), \text { where } f(x)=\left\\{\begin{array}{ll}{x^{3}+1,} & {x<1} \\ {x+1,} & {x \geq 1}\end{array}\right. $$
Using the Intermediate Value Theorem In Exercises \(91-94,\) use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval \([0,1] .\) Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ g(t)=2 \cos t-3 t $$
Making a Function Continuous In Exercises \(61-66,\) find the constant \(a,\) or the constants \(a\) and \(b\) , such the function is continuous on the entire real number line. $$ f(x)=\left\\{\begin{array}{ll}{3 x^{2},} & {x \geq 1} \\ {a x-4,} & {x<1}\end{array}\right. $$
Writing In Exercises \(87-90\) , explain why the function has a zero in the given interval. $$ \begin{array}{ll}{\text { Function }} & {\text { Interval }} \\\ {f(x)=\frac{1}{12} x^{4}-x^{3}+4} & {[1,2]}\end{array} $$
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