What is the range of √ x 5?

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

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What is the range of √ x 5?

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What is the range of √ x 5?

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range+y=\sqrt{x-5}-\sqrt{x+5}

en

What is the range of √ x 5?

Related » Graph » Number Line » Similar » Examples »

What is the range of √ x 5?

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Examples

  • x^{2}-x-6=0
  • -x+3\gt 2x+1
  • line\:(1,\:2),\:(3,\:1)
  • f(x)=x^3
  • prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x)
  • \frac{d}{dx}(\frac{3x+9}{2-x})
  • (\sin^2(\theta))'
  • \sin(120)
  • \lim _{x\to 0}(x\ln (x))
  • \int e^x\cos (x)dx
  • \int_{0}^{\pi}\sin(x)dx
  • \sum_{n=0}^{\infty}\frac{3}{2^n}

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domain f(x)=\sqrt{x-5}

en

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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

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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

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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

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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

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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

What is the range of √ x 5?

Parth ThakkarParth Thakkar

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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,∞).