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Math  /  Algebra

QuestionFor f(x)=2x−4f(x)=2 x-4 and g(x)=2x2−1g(x)=2 x^{2}-1, find the following functions. a. (f∘g)(x);b⋅(g∘f)(x);c⋅(f∘g)(−1);d.(g∘f)(−1)(f \circ g)(x) ; b \cdot(g \circ f)(x) ; c \cdot(f \circ g)(-1) ; d .(g \circ f)(-1) a. (f∘g)(x)=(f \circ g)(x)= □\square (Simplify your answer.)

Studdy Solution

STEP 1

1. The composition of functions involves substituting one function into another.
2. The notation (f∘g)(x)(f \circ g)(x) represents the composition f(g(x))f(g(x)).
3. The functions given are f(x)=2x−4f(x) = 2x - 4 and g(x)=2x2−1g(x) = 2x^2 - 1.
4. We are required to find and simplify the compositions of these functions.

STEP 2

1. Compute (f∘g)(x)(f \circ g)(x).
2. Compute (g∘f)(x)(g \circ f)(x).
3. Evaluate (f∘g)(−1)(f \circ g)(-1).
4. Evaluate (g∘f)(−1)(g \circ f)(-1).

STEP 3

Compute (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)).
First, substitute g(x)g(x) into f(x)f(x): f(g(x))=f(2x2−1)f(g(x)) = f(2x^2 - 1)

STEP 4

Apply the function ff to g(x)=2x2−1g(x) = 2x^2 - 1.
f(2x2−1)=2(2x2−1)−4f(2x^2 - 1) = 2(2x^2 - 1) - 4

STEP 5

Simplify the expression:
f(2x2−1)=4x2−2−4=4x2−6f(2x^2 - 1) = 4x^2 - 2 - 4 = 4x^2 - 6
Therefore: (f∘g)(x)=4x2−6(f \circ g)(x) = 4x^2 - 6

STEP 6

Compute (g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x)).
First, substitute f(x)f(x) into g(x)g(x): g(f(x))=g(2x−4)g(f(x)) = g(2x - 4)

STEP 7

Apply the function gg to f(x)=2x−4f(x) = 2x - 4.
g(2x−4)=2(2x−4)2−1g(2x - 4) = 2(2x - 4)^2 - 1

STEP 8

Simplify the expression inside the function gg.
g(2x−4)=2(4x2−16x+16)−1=8x2−32x+32−1g(2x - 4) = 2(4x^2 - 16x + 16) - 1 = 8x^2 - 32x + 32 - 1
Therefore: (g∘f)(x)=8x2−32x+31(g \circ f)(x) = 8x^2 - 32x + 31

STEP 9

Evaluate (f∘g)(−1)(f \circ g)(-1).
Substitute x=−1x = -1 into (f∘g)(x)=4x2−6(f \circ g)(x) = 4x^2 - 6: (f∘g)(−1)=4(−1)2−6=4−6=−2(f \circ g)(-1) = 4(-1)^2 - 6 = 4 - 6 = -2

STEP 10

Evaluate (g∘f)(−1)(g \circ f)(-1).
Substitute x=−1x = -1 into (g∘f)(x)=8x2−32x+31(g \circ f)(x) = 8x^2 - 32x + 31: (g∘f)(−1)=8(−1)2−32(−1)+31=8+32+31=71(g \circ f)(-1) = 8(-1)^2 - 32(-1) + 31 = 8 + 32 + 31 = 71
Solution:
a. (f∘g)(x)=4x2−6 (f \circ g)(x) = 4x^2 - 6
b. (g∘f)(x)=8x2−32x+31 (g \circ f)(x) = 8x^2 - 32x + 31
c. (f∘g)(−1)=−2 (f \circ g)(-1) = -2
d. (g∘f)(−1)=71 (g \circ f)(-1) = 71

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