Answer:
*Replace each "x" with the required numbers .
1) g (14) ; Ans;
[tex] g(x) = 2x \\ g(14) = 2(14) \\ g(14)= 28 [/tex]
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2) f (-8) ; Ans;
[tex]f(x) = x + 3 \\ f( - 8) = - 8 + 3 \\ f( - 8) = - 5[/tex]
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3) j (3) ; Ans;
[tex]j(x) = ( - 1)^{x + 2} \\ j(3) = {( - 1)}^{3 + 2} \\ j(3) = {( - 1)}^{5} \\ j(3) = - 1[/tex]
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4) h (4) ; Ans;
[tex]h(x) = {3}^{x} \\ h(4) = {3}^{4} \\ h(4) = 81[/tex]
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5) f (2.8) ; Ans;
[tex]f(x) = x + 3 \\ f(2.8) = 2.8 + 3 \\ f(2.8) = 5.8[/tex]
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6) j(-5) ; Ans;
[tex]j(x) = {( - 1)}^{x + 2} \\ j( - 5) = {( - 1)}^{ - 5 + 2} \\ j( - 5) = {( - 1)}^{ - 3} \\ \\ j( - 5) = \frac{1}{ {( - 1)}^{3} } \\ \\ j( - 5) = \frac{1}{ - 1} \\ \\ j( - 5) = - 1[/tex]
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7) g(3/4); Ans;
[tex]g(x) = 2x \\ \\ g( \frac{3}{4} ) = 2( \frac{3}{4} ) \\ \\ g( \frac{3}{4} ) = \frac{3}{2} \\ \\ g( \frac{3}{4} ) = 1.5[/tex]
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8) [h(6.2)]² ; Ans;
[tex]h(x) = {3}^{x} \\ h(6.2) = {3}^{6.2} \\ h(6.2) = 908.13785497 \\ \\ {(h(6.2))}^{2} = ({3}^{6.2})^{2} \\ {(h(6.2))}^{2} = 824714.36364715[/tex]
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9) j(2) + g(-3) ; Ans;
[tex]j(x) = {( - 1)}^{x + 2} \\ j(2) = {( - 1)}^{2 + 2} \\ j(2) = {( - 1)}^{4} \\ j(2) = 1 \\ \\ g(x) = 2x \\ g( - 3) = 2( - 3) \\ g( - 3) = - 6 \\ \\ \\ j(x) + g(x) \\ j(2) + g( - 3) \\ 1 + ( - 6) \\ 1 - 6 \\ j(2) + g( - 3) = - 5[/tex]
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10) h(1) - f(-10) ; Ans;
[tex]h(x) = {3}^{x} \\ h(1) = {3}^{1} \\ h(1) = 3 \\ \\ f(x) = x + 3\\ f( - 10) = - 10 + 3 \\ f( - 10) = - 7 \\ \\ \\ h(x) - f(x) \\ h(1) - f( - 10) \\ 3 - ( - 7) \\ 3 + 7 \\ = 10 \\ h(1) - f( - 10) = 10[/tex]
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**We look at the table. If f(4) or f(-2) is required, it came from f(x) by substitution each x with the required numbers, and the result is in the table f(x) .
***But if he says that f(x) is equal to a number, he means the resulting table f(x) and asks for the "x" that you take from the table "x" .
Table (1) Ans;
f(4) = –3
f(-2) = – 6
If f(x)= -3 ,then x = 4
If f(x) = 1 , then x = 3
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Table (2) Ans;
f(4) = –7
f(-2) = 1
If f(x)= -3 ,then x = 0
If f(x) = 1 , then x = - 2
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Table (3)Ans;
f(4) = 7
f(-2) = 3
If f(x)= -3 ,then x = 6
If f(x) = 1 , then x = 0
I hope I helped you^_^
