Math

QuestionEvaluate the following compositions: (fg)(1)(f \circ g)(1), (fg)(1)(f \circ g)(-1), (gf)(0)(g \circ f)(0), (gf)(1)(g \circ f)(-1), (gg)(2)(g \circ g)(-2), (ff)(1)(f \circ f)(-1). Use values from f(x)f(x) and g(x)g(x): f(3)=6f(-3)=-6, f(2)=4f(-2)=-4, f(1)=2f(-1)=-2, f(0)=1f(0)=-1, f(1)=2f(1)=2, f(2)=4f(2)=4, f(3)=6f(3)=6, g(3)=6g(-3)=6, g(2)=2g(-2)=2, g(1)=0g(-1)=0, g(0)=1g(0)=-1, g(1)=0g(1)=0, g(2)=2g(2)=2, g(3)=6g(3)=6.

Studdy Solution

STEP 1

Assumptions1. We have two functions, f(x) and g(x), with their values given in the table for different x values. . The symbol (fg)(x)(f \circ g)(x) represents the composition of functions f and g, which means f(g(x))f(g(x)).
3. Similarly, (gf)(x)(g \circ f)(x) represents g(f(x))g(f(x)).
4. We are asked to evaluate these compositions for given x values.

STEP 2

Let's start with the first expression, (fg)(1)(f \circ g)(1). This represents f(g(1))f(g(1)).

STEP 3

First, we need to find the value of g(1)g(1). Looking at the table, we see that g(1)=0g(1) =0.

STEP 4

Now, we substitute g(1)g(1) into the function f(x)f(x), so we have f(g(1))=f(0)f(g(1)) = f(0).

STEP 5

Looking at the table, we see that f(0)=1f(0) = -1. So, (fg)(1)=1(f \circ g)(1) = -1.

STEP 6

Next, let's evaluate (gf)(0)(g \circ f)(0). This represents g(f(0))g(f(0)).

STEP 7

First, we need to find the value of f(0)f(0). Looking at the table, we see that f(0)=1f(0) = -1.

STEP 8

Now, we substitute f(0)f(0) into the function g(x)g(x), so we have g(f(0))=g(1)g(f(0)) = g(-1).

STEP 9

Looking at the table, we see that g()=g(-) =. So, (gf)()=(g \circ f)() =.

STEP 10

Next, let's evaluate (fg)()(f \circ g)(-). This represents f(g())f(g(-)).

STEP 11

First, we need to find the value of g()g(-). Looking at the table, we see that g()=0g(-) =0.

STEP 12

Now, we substitute g()g(-) into the function f(x)f(x), so we have f(g())=f(0)f(g(-)) = f(0).

STEP 13

Looking at the table, we see that f(0)=f(0) = -. So, (fg)()=(f \circ g)(-) = -.

STEP 14

Next, let's evaluate (gg)(2)(g \circ g)(-2). This represents g(g(2))g(g(-2)).

STEP 15

First, we need to find the value of g(2)g(-2). Looking at the table, we see that g(2)=2g(-2) =2.

STEP 16

Now, we substitute g(2)g(-2) into the function g(x)g(x), so we have g(g(2))=g(2)g(g(-2)) = g(2).

STEP 17

Looking at the table, we see that g(2)=2g(2) =2. So, (gg)(2)=2(g \circ g)(-2) =2.

STEP 18

Next, let's evaluate (gf)()(g \circ f)(-). This represents g(f())g(f(-)).

STEP 19

First, we need to find the value of f(1)f(-1). Looking at the table, we see that f(1)=f(-1) = -.

STEP 20

Now, we substitute f()f(-) into the function g(x)g(x), so we have g(f())=g()g(f(-)) = g(-).

STEP 21

Looking at the table, we see that g()=g(-) =. So, (gf)(1)=(g \circ f)(-1) =.

STEP 22

Finally, let's evaluate (ff)(1)(f \circ f)(-1). This represents f(f(1))f(f(-1)).

STEP 23

First, we need to find the value of f(1)f(-1). Looking at the table, we see that f(1)=f(-1) = -.

STEP 24

Now, we substitute f(1)f(-1) into the function f(x)f(x), so we have f(f(1))=f()f(f(-1)) = f(-).

STEP 25

Looking at the table, we see that f()=4f(-) = -4. So, (ff)(1)=4(f \circ f)(-1) = -4.
So, the solutions area. (fg)(1)=1(f \circ g)(1) = -1 d. (gf)(0)=0(g \circ f)(0) =0 b. (fg)(1)=1(f \circ g)(-1) = -1 e. (gg)()=(g \circ g)(-) = c. (gf)(1)=(g \circ f)(-1) = f. (ff)(1)=4(f \circ f)(-1) = -4

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