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Determine all values of the constant a such that the following function is continuous for all real numbers. f(x) = {ax/tan x, x ≥ 0 {a²-2, x < 0


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We have : f(x)={axtanx for x0a22 for x<0Recall that for a function to be continuous at a point c it must be true that : limxcf(x)=limxc+f(x)=f(c)Now the functions f(x)=axtanxg(x)=a22Are continuous at the intervals between their asymptotes and the other function is a constant,meaning that the only point of interest is when the domains merge at x=0When approaching the value x0 from the right the function is : f(x)=axtanxSo the limit from the right is limx0+axtanx=00This value is undefined, however the quotient property defined at page 49 property 4, can be tweaked (case when L = 1) to write if :limxcf(x)=KThen it is also true that :limxc1f(x)=1KSo let’s calculate this :limx0+1f(x)=limx0+tanxax=limx0+sinxaxcosxRecall and use the product of limits property page 49 ,3 :limx0+1f(x)=limx0+sinxxlimx0+1acosxlimx0+1f(x)=1alimx0+sinxxRecall Theorem 1.9 case 1 page 65 that states :limx0sinxxUse it in our case : limx0+1f(x)=1aBecause of what we showed above it also means that : limx0+f(x)=aWhen approaching the value x0 from the left the function is f(x)=a22So the limit from the right is limx0a22=a22If the condition of continuity is to be satisfied then : limx0a22=limx0+axtanxSo : a22=aa2a2=0Solve it: D=(1)24(2)(1)=9a1,2=1±92Write the solution: a{1,2}For these two value of a the given f(x) is continuous for every real number \begin{gather*} \color{Mahogany} \text {We have : }\\ f(x) = \begin{cases} \dfrac{ax}{\tan x} \hspace{5mm} \text{ for }\hspace{5mm} x\geq 0 \\ a^2 - 2 \hspace{5mm} \text{ for }\hspace{5mm} x< 0 \\ \end{cases}\\\\ \color{Mahogany} \text {Recall that for a function to be continuous at a point c it must be true that : }\\ \lim\limits _{x\to c^{-}}f(x) = \lim\limits _{x\to c^{+}}f(x) = f(c)\\\\ \color{Mahogany} \text {Now the functions }\\ f(x) =\frac{ax}{\tan x} \\ g(x) = a^2 - 2\\\\ \color{Mahogany} \text {Are continuous at the intervals between their asymptotes and the other function }\\ \color{Mahogany} \text {is a constant,meaning that the only point of interest is when the domains merge }\\ \color{Mahogany} \text {at $x = 0$}\\ \color{Mahogany} \text {When approaching the value $x \to 0 $ from the right the function is : }\\ f(x) = \frac{ax}{\tan x} \\\\ \color{Mahogany} \text {So the limit from the right is }\\ \lim\limits_{x\to 0^{+}} \frac{ax}{\tan x} = \frac{0}{0}\\\\ \color{Mahogany} \text {This value is undefined, however the quotient property defined at page }\\ \color{Mahogany} \text {49 property 4, can be tweaked (case when L = 1) to write if :}\\ \lim\limits_{x\to c} f(x) = K\\\\ \color{Mahogany} \text {Then it is also true that :}\\ \lim\limits_{x\to c} \frac{1}{f(x)} = \frac{1}{K}\\\\ \color{Mahogany} \text {So let's calculate this :}\\ \begin{align*} \lim\limits_{x\to 0^{+}} \frac{1}{f(x)} & = \lim\limits_{x\to 0^{+}} \frac{\tan x}{a x} \\ & = \lim\limits_{x\to 0^{+}} \frac{\sin x}{a x \cos x} \\ \end{align*}\\ \color{Mahogany} \text {Recall and use the product of limits property page 49 ,3 :}\\ \lim\limits_{x\to 0^{+}} \frac{1}{f(x)} = \lim\limits_{x\to 0^{+}} \frac{\sin x}{ x} \lim\limits_{x\to 0^{+}} \frac{1}{a\cos x}\\ \lim\limits_{x\to 0^{+}} \frac{1}{f(x)} = \frac{1}{a}\lim\limits_{x\to 0^{+}} \frac{\sin x}{ x}\\\\ \color{Mahogany} \text {Recall Theorem 1.9 case 1 page 65 that states :}\\ \lim\limits_{x\to 0} \frac{\sin x}{ x} \\\\ \color{Mahogany} \text {Use it in our case :}\\\ \lim\limits_{x\to 0^{+}} \frac{1}{f(x)} = \frac{1}{a}\\\\ \color{Mahogany} \text {Because of what we showed above it also means that :}\\\ \lim\limits_{x\to 0^{+}} f(x) = a\\\\ \color{Mahogany} \text {When approaching the value $x \to 0 $ from the left the function is }\\ f(x) = a^2 - 2 \\\\ \color{Mahogany} \text {So the limit from the right is }\\ \lim\limits_{x\to 0^{-}}a^2 - 2 = a^2 - 2 \\\\ \color{Mahogany} \text {If the condition of continuity is to be satisfied then : }\\ \lim\limits_{x\to 0^{-}} a^2 - 2 = \lim\limits_{x\to 0^{+}}\frac{ax}{\tan x} \\\\ \color{Mahogany} \text {So : }\\ a^2 - 2 = a \\ a^2 -a - 2 =0\\ \color{Mahogany} \text {Solve it: }\\ D = (-1)^2 -4(-2)(1) = 9\\ a_{1,2} = \frac{1\pm \sqrt{9}}{2}\\\\ \color{Mahogany} \text {Write the solution: }\\ a \in \{-1,2\}\\\\ \color{Mahogany} \text {For these two value of a the given f(x) is continuous for every real number }\\ \end{gather*}

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