PROBLEM
Suppose that the functions s and t are defined for all real numbers x as follows.
s(x)=4xt(x)=3x−1 Write the expressions for (s⋅t)(x) and (s+t)(x) and evaluate (s−t)(−2).
(s⋅t)(x)=(s+t)(x)=(s−t)(−2)= □
□
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STEP 1
1. We are given the functions s(x)=4x and t(x)=3x−1.
2. We need to find expressions for (s⋅t)(x) and (s+t)(x).
3. We need to evaluate (s−t)(−2).
STEP 2
1. Find the expression for (s⋅t)(x).
2. Find the expression for (s+t)(x).
3. Evaluate (s−t)(−2).
STEP 3
To find (s⋅t)(x), multiply the functions s(x) and t(x):
(s⋅t)(x)=s(x)⋅t(x)=(4x)⋅(3x−1)
STEP 4
Distribute to simplify the expression:
(s⋅t)(x)=4x⋅3x−4x⋅1 (s⋅t)(x)=12x2−4x
STEP 5
To find (s+t)(x), add the functions s(x) and t(x):
(s+t)(x)=s(x)+t(x)=4x+(3x−1)
STEP 6
Simplify the expression:
(s+t)(x)=4x+3x−1 (s+t)(x)=7x−1
STEP 7
To evaluate (s−t)(−2), first find the expression for (s−t)(x):
(s−t)(x)=s(x)−t(x)=4x−(3x−1)
STEP 8
Simplify the expression:
(s−t)(x)=4x−3x+1 (s−t)(x)=x+1
SOLUTION
Substitute x=−2 into (s−t)(x):
(s−t)(−2)=(−2)+1 (s−t)(−2)=−1 The expressions and evaluation are:
(s⋅t)(x)=12x2−4x(s+t)(x)=7x−1(s−t)(−2)=−1
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