Question

What is range of f(x)= sgn(2^x)+sgn(|x-5|)
Please solve it with full solution i am really in trouble

Answer 1

Range and Signum Function

Let's first get a clear definition of Range as well as the Signum Function.

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Range of a Function

A function $f$ defined as $f: A \to B$ has a Domain $A$ and a Co-Domain $B$

The subset of this co-domain $B$ which represents the set of all possible outputs of function $f$ is known as the Range.

In other words, list out all possible outputs that a function can have, and put them in a Set. This Set is the Range.

The range of a function can be a discrete set consisting of only a few values, or it may be an infinite continuous set like $\mathbb{R}$ (the set of Real Numbers) or some discrete value set with infinite elements, like $\mathbb{N}$. The concept is the same. The output is called Range.

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Signum Function

The Signum Function (also called the Sign Function) has a very particular definition:

$sgn(f(x)) = \begin{cases} -1 & \text{if }f(x)<0 \\ 0 & \text{if }f(x)=0\\ 1 & \text{if }f(x)>0\end{cases}$

The output is quite simple. 1 if function value is positive. -1 if function value is negative and 0 if function value is zero.

The Domain is the entire set of Real Numbers $\mathbb{R}$. Which means f(x) can take any real value.

The Range of Signum Function is the Set {-1, 0, 1}, as it has only three possible outputs.

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The Answer

Let's come to the question in hand.

$f(x) = sgn(2^x)+sgn(|x-5|)$

The function has the sum of two parts. Let's look at each part individually.

• $sgn(2^x)$

$2^x$ is an Exponential function and is always positive. It never takes a negative or zero value.

Since $2^x > 0$ always, we have $sgn(2^x)=1$ for all values of $x\in \mathbb{R}$.

• $sgn(|x-5|)$

$|x-5|$ is a Modulus. Absolute value. Which means it is always non-negative.

$|x-5| > 0\quad \forall\ x\neq 5,\ x\in \mathbb{R} \\\\ |x-5| = 0\quad \text{for } x=5$

So, essentially, when we apply the signum function, we have:

$sgn(|x-5|) = \begin{cases} 1 & \forall\ x\neq 5,\ x\in \mathbb{R} \\ 0 & \text{for }x=5\end{cases}$

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When we combine the two parts of the function, we see that:

$f(x) = \begin{cases} 1+1 = 2 & \text{for }x\neq 5 \\ 1+0 = 1 &\text{for }x=5 \end{cases}$

Hence, the function f(x) has only two possible outputs.

So, the Range is {1, 2}

Answer 2

Answer:

The correct answer is the Range is therefore 1, 2,

Explanation:

It's the question of Range and Signum Function.

$f(x)=sgn(2^{x} ) + sgn(|x-5|)$

Two components make up the function's sum. Let's examine each component separately.

$sgn(2^{x} )$

$2^{x}$ is an exponential function that can never be negative. Never is a negative or zero value accepted.

As, $2^{x} > 0$ , we have had $sgn(2^{x} ) = 1$ for every value of x ∈ R.

$|x-5|$ the Modulus. absolute worth. Which implies that it is never negative.

$|x-5|$ > 0 ∀ x ≠ 5, x ∈ R

$|x-5|$ = 0 for $x=5$

Therefore, when we use the signum function, we essentially have:

$sgn(|x-5|)=\left \{ {{1, x is not equal to 5} \atop {0, x=5}} \right.$

Combining the two components of the function, we note that:

$f(x) = \left \{ {{1+1=2 for x is not equal to 5} \atop {1+0=1 for x=5} \right.$

Thus, there are only two possible outputs for the function f(x).

The Range is therefore 1, 2.

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