hello everyone here the question take that find the domain and range of a function f x is equal to 1 by x minus 5 years we have even the function f x is equal to effects is equal to 15 square minus 5 square root of any number is greater than zero 2 x minus 5 is greater than 28 also Meenu mean of effects library infinity now I have to find it Show range it lies between Light Between Infinity 15 2 x minus y life between infinity 15 what is -5 life between greater than 1.0 and smaller than 15 infinity and 150 is equal to infinitive 215 square root of x minus 5 wallop and infinitive but range of FX shayari wallpaper real number positive this is the ringtone thank you Our online expert tutors can answer this problem Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Your first 5 questions are on us! You are being redirected to Course Hero I want to submit the same problem to Course HeroCorrect Answer :) Let's Try Again :( Try to further simplify Number LineGraphHide Plot » Sorry, your browser does not support this applicationExamples
step-by-step range+y=\sqrt{x-5}-\sqrt{x+5} en Our online expert tutors can answer this problem Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Your first 5 questions are on us! You are being redirected to Course Hero I want to submit the same problem to Course HeroCorrect Answer :) Let's Try Again :( Try to further simplify Number LineGraphHide Plot » Sorry, your browser does not support this applicationExamples
step-by-step domain f(x)=\sqrt{x-5} en $\begingroup$ What's the domain of $\ f(x) = \sqrt{\frac{x - 3}{x - 5}} $ ? My guess is there are two possibilities depending on whether $ x - 5 $ is positive or negative, after excluding $ 5 $ of course. asked Nov 2, 2013 at 9:43
$\endgroup$ $\begingroup$ The term inside the square root has to be positive.Also $x\neq5$ as the denominator then becomes 0. Hence, $$\frac{x-3}{x-5}\ge0$$ $$\Rightarrow x\in (\infty,3]\cup(5,\infty)$$ answered Nov 2, 2013 at 10:03
GTX OCGTX OC 1,5292 gold badges12 silver badges19 bronze badges $\endgroup$ 0 $\begingroup$ Assuming a function from $\mathbb{R}\to\mathbb{R}$, $\displaystyle{\frac{x-3}{x-5}}$ must be non-negative, so $x-3 \geq0$ and $x-5\geq0 \implies$ $x\geq5$. Also, $x-3\leq0$ and $x-5\leq0 \implies x\leq3$. Also, $x\neq5$ or we'll be dividing by zero. So the domain is $\{x\in\mathbb{R}:x\leq3\;\text{or}\;x>5\}$. If it's the complex square root function, then the domain is simply $\{x\in\mathbb{R}:x\neq5\}$. answered Nov 2, 2013 at 10:48
$\endgroup$ $\begingroup$ You seem to know that the inside of the square root must be non-negative. In your case, (1) $\dfrac{x-3}{x-5} \geq 0$. Moreover, as you've found, the denominator can not be zero. Namely (2) $x-5 \ne 0$. Under the condition, let's consider the condition (1). $$\frac{x-3}{x-5} \geq 0 \iff \frac{x-3}{x-5}(x-5)^2 \geq 0.$$ For the equivalence, I used the fact that the square of a real number is always non-negative. I think that you can simplify the LHS of the last inequality and reach the answer. answered Nov 2, 2013 at 11:04
H. ShindohH. Shindoh 1,98015 silver badges24 bronze badges $\endgroup$ 4 $\begingroup$ HINT: The thing under the root must be positive. That happens when both numerator and denominator are positive, or when both are negative. answered Nov 2, 2013 at 9:45
Parth ThakkarParth Thakkar 4,2543 gold badges29 silver badges49 bronze badges $\endgroup$ What is the range of the function f X √ X 5?The square root function never produces a negative result. Therefore, for the function f(x)=√x+5 , the domain is {x∈R∣x≥−5} and the range is {f(x)∈R∣f(x)≥0} .
What is the range of √ X?We know that the range of the square root function √ 𝑥 is [ 0 , ∞ [ . In other words, for any number 𝑦 in the interval [ 0 , ∞ [ , we can find some number 𝑥 that satisfies 𝑦 = √ 𝑥 .
What is the range of √?Thus, the range of the square root function is [0,∞).
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