$6$ . Diketahui fungsi $g$ : sumbu $\mathrm{X} \rightarrow \mathrm{V}$ yang didefinisikan sebagai berikut : Apabila $P(x, 0)$ maka $g(P)=\left(x, x^{2}\right)$ . $\newline$ a). Tentukan peta $A(3,0)$ oleh $g$ $\newline$ b) Apakah $R(-14,196) \in$ daerah nilai $g$ ? $\newline$ c) Apakah g surjektif ? $\newline$ d) Gambarlah daerah nilai $g$

Full solution

6

. Diketahui fungsi

g

: sumbu

\mathrm{X} \rightarrow \mathrm{V}

yang didefinisikan sebagai berikut : Apabila

P(x, 0)

maka

g(P)=\left(x, x^{2}\right)

\newline

a). Tentukan peta

A(3,0)

oleh

g

\newline

b) Apakah

R(-14,196) \in

daerah nilai

g

\newline

c) Apakah g surjektif ?

\newline

d) Gambarlah daerah nilai

g

Substitute and Calculate: To find the image of $A(3,0)$ under $g$ , we substitute $x$ with $3$ in $g(P)=(x,x^{2})$ . $\newline$ $g(A) = g(3,0) = (3,3^{2}) = (3,9)$ .
Check for Range Membership: To check if $R(-14,196)$ is in the range of $g$ , we look for an $x$ such that $g(P)=(x,x^{2})$ equals $(-14,196)$ . Since the second component is $x^{2}$ , we take the square root of $196$ to find $x$ . $\sqrt{196} = 14$ , but we need $-14$ , so $R(-14,196)$ is not in the range because we can't get a negative number from squaring $x$ .
Determine Surjectivity: To determine if $g$ is surjective, we need to check if every element in the codomain $V$ has a preimage in the domain $X$ . Since $g(P)=(x,x^{2})$ , $g$ can only produce non-negative second components because squares are non-negative. Therefore, $g$ is not surjective as it does not cover negative numbers in the second component of the codomain $V$ .
Draw Range of g: To draw the range of g, we plot the points $(x,x^{2})$ for all $x$ in the domain. $\newline$ This is a parabola opening upwards with the vertex at the origin $(0,0)$ .