Function should remain real so math√(x^2–1)/mathshould not be equal to zero so math x^2–1 \neq 0/math mathx^2=1/math Or mathx=\pm 1/math AndThe domain and range of the function f= ((1/1x2)) x ∈ R, x ≠ ± 1 are respectively (A) R 1, 1 (∞, 0) ∪ 1, ∞) (B) R, (∞, 0) ∪ 1 f (x) = 1/√x−5 Now for real value of x5≠0 and x5>0 ⇒ x≠5 and x>5 Hence the domain of f = (5, ∞) And the range of a function consists of all the second elements of all the ordered pairs, ie, f(x), so we have to find the values of f(x) to get the required range Now we know for this function x5>0 taking square root on both
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F x x 1 -3 domain and range-To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Find domain and range of `f(x)=x/(1x^2)`F(x)=1 if x4>0 f(x)=1 if x4
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SOLUTION find the range and domain f (x)=1/x2 Question find the range and domain You can put this solution on YOUR website!The "" means "such that," the symbol ∈ means "element of," and "ℝ" means "all real numbers" Putting it all together, this statement can be read as "the domain is the set of all x such that x is an element of all real numbers" The range of f (x) = x2 in set notation is R {y y ≥ 0} R indicates rangeFind the Domain and Range f (x)= (x1)/ (x1) f (x) = x − 1 x 1 f ( x) = x 1 x 1 Set the denominator in x−1 x1 x 1 x 1 equal to 0 0 to find where the expression is undefined x1 = 0 x 1 = 0 Subtract 1 1 from both sides of the equation x = −1 x = 1
Algebra Find the Domain and Range f (x)=x^21 f (x) = x2 1 f ( x) = x 2 1 The domain of the expression is all real numbers except where the expression is undefined In this case, there is no real number that makes the expression undefined Interval NotationQuestion 1 Find the domain and range of the following functions f(x) = x 3 Solution Domain A set of all defined values of x is known as domain Range The out comes or values that we get for y is known as range Domain for given function f(x) = x 3 For any real values of x, f(x) will give defined values Hence the domain is R Misc 4 Find the domain and the range of the real function f defined by f(x) = √((𝑥−1)) It is given that the function is a real function Hence, both its domain and range should be real numbers x can be a number greater 1 Here, f(x) is always positive, Minimum value of f(x) is 0,
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The domain and range are the intervals in which the function is defined in the x and y axes Answer and Explanation 1 We are given the function {eq}f(x)=2x^23x1 {/eq}(0, infinity) , {x/x > 0) for any integer Range;Find the Domain and Range f (x)=1 f (x) = 1 f ( x) = 1 The domain of the expression is all real numbers except where the expression is undefined In this case, there is no real number that makes the expression undefined Interval Notation (−∞,∞) ( ∞, ∞) Set Builder Notation {xx ∈ R} { x x
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Domain is all reals except 1 (Can't divide by 0) Range if y=1/(x1), then y(x1)=1 son xyy=1 and xy=1y, finally, then x=(1y)/y y can take any value except 0 Range is all reals except 0 graph{y=1/(x1) 974, 803, 38, 509}Find Domain and Range of real functions (1) `f(x)=(x2)/(3x)` (2)`f(x)=1/sqrt(x5)` (3) `f(x)=x/(1x^2)`Arrow_forward Question View transcribed image text fullscreen Expand check_circle Expert Answer Want to see the stepbystep answer?
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Were given a function and we're asked to find the domain and range of dysfunction Function is F of X equals one plus X square Notice that the function F is a polynomial function that's a quadratic function and therefore the domain of F is all real numbersActually, at first glance, I don't know So I have to break this down into parts that I do know tan(x) = sin(x) / cos(x) And I know 1/tan(x) = cos(x)/sin(x), so to find the domain I have to look for the places where the denominator is zero (first114 Range of a function For a function f X → Y the range of f is the set of yvalues such that y = f(x) for some x in X This corresponds to the set of yvalues when we describe a function as a set of ordered pairs (x,y) The function y = √ x has range;
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Transcript Misc 6 Let f = {("x, " 𝑥2/(1𝑥2)) x ∈ R } be a function from R into R Determine the range of f f = { ("x , " 𝑥2/(1𝑥2)) x ∈ R } We find different values of 𝑥2/(1 𝑥2) for different values of x Domain Value will always be between 0 & 1 We note that Value of range y = 𝑥2/(1 𝑥2) is always positive Also, it is always between 0 and 1 Hence, Range is 1 Confirm that you have a quadratic function A quadratic function has the form ax 2 bx c f (x) = 2x 2 3x 4 The shape of a quadratic function on a graph is parabola pointing up or down There are different methods to calculating the range of a function depending on the type you are working withAnswer and Explanation 1 Given f(x) = 1 x f ( x) = 1 x Domain Since the function is a fraction, the value of the variable are all real numbers except 0 0 since it will result to an undefined
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