Question

For $h\left( x \right)=19;$ how do you find $h\left( 4 \right),h\left( -6 \right)$ and $h\left( 12 \right)?$

Answer 1

Given is a constant function. A constant function never changes its value. The image set of a constant function will be a singleton set that contains the constant image value.

Complete step by step solution:
Consider the given function $h$ defined as $h\left( x \right)=19.$
Because the left-hand side of the definition of the equation constitutes only a single constant value, this function is called a constant function that never changes its image value.
That is, the image value of the constant function is always the same at any point of the domain of the function.
And thus, the image set or the range set will be a singleton set that contains only one image.
Since $h\left( x \right)=19$ is a constant function and the image set or range set is \left\\{ 19 \right\\}, we can say the value of the function at any point is $19.$
That means, the value of the function at a point $4$ from the domain of the function is $19.$
That is, $h\left( 4 \right)=19.$
Also, the value of the function at a point $-6$ from the domain of the function is $19.$
We write, $h\left( -6 \right)=19.$
Similarly, the value of the function at a point $12$ from the domain of the function is $19.$
Again, $h\left( 12 \right)=19.$

Hence, we get $h\left( x \right)=h\left( 4 \right)=h\left( -6 \right)=h\left( 12 \right)=19.$

Note: A constant function is onto. Because, every element in the codomain has a preimage in the domain. But a constant function is not ono-one. The above given function is an example of this fact. Because, every point is mapped to a single point.