Question 1204639
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Part (a)


f(x) = -2x+4
f(3) = -2(3)+4
f(3) = -6+4
f(3) = <font color=red size=4>-2</font>


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Part (b)


g(x) = x^2-3x
g(-2) = (-2)^2-3(-2)
g(-2) = 4+6
g(-2) = <font color=red size=4>10</font>


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Part (c)


g(f(3)) = g(-2) because f(3) = -2
We replace f(3) with -2


Then refer to part (b) to find that g(-2) = <font color=red size=4>10</font> which must mean g(f(3)) = <font color=red size=4>10</font>


Here's another approach
g(x) = x^2-3x
g(x) = (x)^2-3(x)
g(f(x)) = ( f(x) )^2-3( f(x) )
g(f(x)) = (-2x+4)^2-3(-2x+4)
g(f(x)) = (4x^2-16x+16)+(6x-12)
g(f(x)) = 4x^2+(-16x+6x)+(16-12)
g(f(x)) = 4x^2 - 10x + 4


Then,
g(f(x)) = 4x^2 - 10x + 4
g(f(3)) = 4(3)^2 - 10(3) + 4
g(f(3)) = 4(9) - 10(3) + 4
g(f(3)) = 36 - 30 + 4
g(f(3)) = 6 + 4
g(f(3)) = <font color=red size=4>10</font>
Your steps do not need to be as verbose as shown above.


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Part (d)


f(x) = -2x + 4
f(x) = -2( x ) + 4
f(g(x)) = -2( g(x) ) + 4
f(g(x)) = -2( x^2-3x ) + 4
f(g(x)) = <font color=red size=4>-2x^2 + 6x + 4 </font>


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Summary of the four answers<ol type="a"><li>f(3) = <font color=red size=4>-2</font></li><li>g(-2) = <font color=red size=4>10</font></li><li>g(f(3)) = <font color=red size=4>10</font></li><li>f(g(x)) = <font color=red size=4>-2x^2 + 6x + 4 </font></li></ol>Each answer can be verified with GeoGebra.
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