Question

Range of f(x) = sgn (x2 + 1) is


{1}


{0}


[1, ∞)


{0, 1}

Answer 1

Answer:

hope it will help you

Step-by-step explanation:

Step-by-step explanation: If x belongs to R(all real numbers), then x-2 also belongs to R. Thus, the range of f(x) is the range of sgn(signum function). The range of sgn is {-1,0,1}.

Answer 2

Answer:

Range of $f(x) = sgn (x^{2} + 1)$ is 1.

Step-by-step explanation:

Given: $f(x) = sgn (x^{2} + 1)$

Find: Range of given function is-

(a) {1}

(b) {0}

(c) [1, ∞)

(d) {0, 1}

The sets of all the x-coordinates and all the y-coordinates of ordered pairs, respectively, are the domain and range of a relation. For example, if the relation is$\left \{ R = (1, 2), ( 2, 2), ( 3, ), ( 4, 3) \right \}$:

Domain is the set of all x-coordinates with values of $\left \{ 1, 2, 3, 4 \right \}$

The set of all y-coordinates in the range is equal to $\left \{ 2, 3 \right \}$.

Signum Function: The signum function simply gives the sign for the given values of x. For x value greater than zero, the value of the output is +1, for x value lesser than zero, the value of the output is -1, and for x value equal to zero, the output is equal to zero.

$sgn(x)= 1$ for $x > 0$

$sgn(x)=0$ for $x=0$

$sgn(x)=-1$ for $x < 0$

So,$sgn(x)=\frac{x}{|x|}$

Therefore, $sgn(x^{2} +1)=\frac{x^{2} +1}{|x^{2} +1|}$

If $x$ is real then $x^{2} +1 > 0$

Hence, $sgn(x^{2} +1)=1.$

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