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
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
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range+y=\sqrt{x-5}-\sqrt{x+5}
en
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Number Line
Graph
Hide Plot »
Sorry, your browser does not support this applicationExamples
- 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}
step-by-step
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
Parth ThakkarParth Thakkar
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