Question

If $f(x - 7) = x^2; x \in \{-1, 2, 5\}$ and $g(x) = f(x^2)$ then find domain, range, graph (DRG) of $g(x)$.

Answer 1

Domain of $g(x)$: $\varnothing$

Range of $g(x)$: $\varnothing$

Graph of $g(x)$: Empty graph (no points)

Answer 2

Solution:

We are given that

$$f(x-7)=x^2 \quad \text{for } x\in\{-1,2,5\}.$$

For each $x$ in this set, we have:

• For $x=-1$: $f(-1-7)=f(-8)=(-1)^2=1.$
• For $x=2$: $f(2-7)=f(-5)=2^2=4.$
• For $x=5$: $f(5-7)=f(-2)=5^2=25.$

Thus, the function $f$ is defined only at the points $-8$, $-5$, and $-2$ with:

$$f(-8)=1,\quad f(-5)=4,\quad f(-2)=25.$$

Now, the function $g$ is defined as

$$g(x)=f(x^2).$$

For $g(x)$ to be defined, the input to $f$ must belong to the domain of $f$; that is, we require

$$x^2 \in \{-8,-5,-2\}.$$

However, note that $x^2$ is always nonnegative for real $x$. Since all elements in the set $\{-8,-5,-2\}$ are negative, there is no real number $x$ such that $x^2 = -8$, $-5$, or $-2$.

Therefore:

• Domain of $g(x)$: $\varnothing$ (the empty set).
• Range of $g(x)$: $\varnothing$ (since no $x$ leads to a defined $g(x)$).
• Graph of $g(x)$: There are no points to plot (empty graph).

Explanation (minimal):

Given $f(x-7)=x^2$ for $x\in\{-1,2,5\}$, we find that $f$ is defined only at $-8$, $-5$, $-2$. For $g(x)=f(x^2)$ to be defined, we require $x^2$ to be one of these negative numbers, which is impossible for real $x$. Hence, $g$ is not defined for any real $x$.