To find the value of q, we need to find d(-8). Put another way, we need to find the value of d(x) when x = -8
[tex]d(x) = -\sqrt{\frac{1}{2}x+4}[/tex]
[tex]d(-8) = -\sqrt{\frac{1}{2}(-8)+4}[/tex]
[tex]d(-8) = -\sqrt{-4+4}[/tex]
[tex]d(-8) = -\sqrt{0}[/tex]
[tex]d(-8) = 0[/tex]
So this means q = 0. Note that -0 is just 0.
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The value of r will be a similar, but now we use f(x) this time.
Plug in x = 0
[tex]f(x) = \sqrt{\frac{1}{2}x+4}[/tex]
[tex]f(0) = \sqrt{\frac{1}{2}*0+4}[/tex]
[tex]f(0) = \sqrt{0+4}[/tex]
[tex]f(0) = \sqrt{4}[/tex]
[tex]f(0) = 2[/tex]
Therefore, r = 2.
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For s, we plug x = 10 into f(x)
[tex]f(x) = \sqrt{\frac{1}{2}x+4}[/tex]
[tex]f(10) = \sqrt{\frac{1}{2}*10+4}[/tex]
[tex]f(10) = \sqrt{5+4}[/tex]
[tex]f(10) = \sqrt{9}[/tex]
[tex]f(10) = 3[/tex]
So s = 3.
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Finally, plug x = 10 into d(x) to find the value of t
[tex]d(x) = -\sqrt{\frac{1}{2}x+4}[/tex]
[tex]d(10) = -\sqrt{\frac{1}{2}(10)+4}[/tex]
[tex]d(10) = -\sqrt{5+4}[/tex]
[tex]d(10) = -\sqrt{9}[/tex]
[tex]d(10) = -3[/tex]
A shortcut you could have taken is to note how d(x) = -f(x), so this means
d(10) = -f(10) = -9 since f(10) = 9 was found in the previous section above.
Whichever method you use, you should find that t = -3.
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