A role \$g\$ is defined as \$g(x)=|4-x|\$, uncover the variety of \$g\$ if its domain is \$-3leq x leq 6\$.

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My attempt:

I substitute \$-3\$ and also \$6\$ right into the function and I got \$2\$ and also \$7\$. And also the prize is between \$0\$ come \$7\$.

What need to I execute to acquire the correct answer ? thanks in advance.

Let"s begin with the graph that the pure value function \$f(x) = |x|\$. Due to the fact that \$|x|\$ represents the street of the number \$x\$ from \$0\$, \$\$|x| = egincasesx && extif \$x geq 0\$\-x && ext{if \$x

Observe the \$|x| = |-x|\$ because the point out \$x\$ and also \$-x\$ are equidistant zero. Hence, \$\$|4 - x| = |-(4 - x)| = |-4 + x| = |x - 4|\$\$The graph the \$y = |x - 4|\$ is obtained from the graph that \$f(x)\$ by translating the graph of \$f(x)\$ by 4 units to the right. You should have the ability to convince yourself of this by make a table of values for \$y = |x|\$ and also \$y = |x - 4|\$. Another way to check out this is to write the piecewise an interpretation
shn stated in the comments\$\$|4 - x| = |x - 4| = egincasesx - 4 && extif \$x geq 4\$\4 - x && ext{if \$x

Restricting the domain \$y = |x - 4|\$ come \$<-3, 6>\$ yields the graph the \$g(x) = |x - 4|, -3 leq x leq 6\$.

From that graph, we check out that \$g\$ has minimum value \$0\$ at \$x = 4\$ (its vertex) and maximum value \$7\$ at \$x = -3\$. Due to the fact that the role is continuous, that assumes every value in between \$0\$ and \$7\$ in the expression \$<-3, 4>\$ and, consequently, in the expression \$<-3, 6>\$.

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