Mathematics, 07.07.2019 12:30 talanna394

Find the interval on which $f(x)=\sqrt{x^{2}-9 }$ is continuous. specify right or left continuity at the endpoints.

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Answer from: tejasheree

Answer

$f$ is continuous on its domain of $(-\infty, -3] \cup[3, \infty)$ .

$f$ is continuous from the left at -1 and continuous from the right at 1.

explanation

assuming domain and codomain of all real numbers, function $f$ is only defined for nonnegative radicands:

$\begin{aligned}x^2 - 9\ge 0 & \implies x^2 \ge 9\\& \implies \sqrt{x^2} \ge \sqrt{9} \\& \implies |x| \ge 3\end{aligned}$

there are many ways of solving a quadratic inequality. here we will reason by cases, using the definition of absolute value.

if $x \ge 0$ then $|x| = x$ and so the inequality turns into $x \ge 3$ when x ≥ 0. of course, "x≥3 when x≥0" simplifies to x ≥ 3.

or, if $x < 0$ , then $|x| = -x$ and so $|x| \ge 3 \implies -x \ge 3 \implies x \le -3$ when $x < 0$ . of course, "x≤-3 when x < 0" simplifies to x ≤ -3.

so the domain of $f$ is $(-\infty, -3] \cup[3, \infty)$

now, $f(x) = \sqrt{x^2-9}$ is a continuous function since it is a composition of continuous functions. $x^2-9$ is a polynomial and $\sqrt{x}$ (the parent) is a root function; both of them are continuous on their domains. therefore, $f$ is continuous on its domain of $(-\infty, -3] \cup[3, \infty)$ .

by definition, a function $f$ is continuous from the left at a number a if $\lim_{x \to a^-}f(x) = f(a)$ and continuous from the right at a number a if $\lim_{x \to a^+}f(x) = f(a)$ .

checking continuity at the end points, note that

$\displaystyle\lim_{x\to -3^-} f(x) = \sqrt{(-3)^2 - 9} = 0 = f(-3)$

and

$\displaystyle\lim_{x\to 3^+} f(x) = \sqrt{(3)^2 - 9} = 0 = f(3)$

so $f$ is continuous from the left at -1 and continuous from the right at 1.

Answer from: Quest

Two because two of one thing is two.

Answer from: Quest

{x = -1/2 , y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer: {x = -1/2 , y = -1

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Find the interval on which [tex]f(x)=\sqrt{x^{2}-9 }[/tex] is continuous. specify right or left cont...