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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