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Sagot :
Sure, let's solve the problem step by step.
We are given two functions: [tex]\( f(x) = 3x + 4 \)[/tex] and [tex]\( g(x) = px + q \)[/tex], where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are constants.
We need to determine the values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex] given the conditions: [tex]\( g(f(-1)) = 1 \)[/tex] and [tex]\( g^2(2) = 5 \)[/tex].
### Step 1: Calculate [tex]\( f(-1) \)[/tex]
First, we find the value of [tex]\( f(-1) \)[/tex]:
[tex]\[ f(x) = 3x + 4 \][/tex]
Substitute [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3(-1) + 4 = -3 + 4 = 1 \][/tex]
### Step 2: Calculate [tex]\( g(f(-1)) \)[/tex]
We now calculate [tex]\( g(f(-1)) \)[/tex]:
[tex]\[ f(-1) = 1 \][/tex]
Therefore:
[tex]\[ g(f(-1)) = g(1) = p(1) + q = p + q \][/tex]
Given [tex]\( g(f(-1)) = 1 \)[/tex]:
[tex]\[ p + q = 1 \quad \text{(Equation 1)} \][/tex]
### Step 3: Calculate [tex]\( g^2(2) \)[/tex]
Next, we calculate [tex]\( g^2(2) \)[/tex] which means [tex]\( g(g(2)) \)[/tex]:
[tex]\[ g(x) = px + q \][/tex]
First, find [tex]\( g(2) \)[/tex]:
[tex]\[ g(2) = p(2) + q = 2p + q \][/tex]
Then calculate [tex]\( g(g(2)) \)[/tex]:
[tex]\[ g(g(2)) = g(2p + q) = p(2p + q) + q = 2p^2 + pq + q \][/tex]
Given [tex]\( g^2(2) = 5 \)[/tex]:
[tex]\[ 2p^2 + pq + q = 5 \quad \text{(Equation 2)} \][/tex]
### Step 4: Solve the system of equations
We have the following system of equations:
[tex]\[ \begin{cases} p + q = 1 & \text{(Equation 1)} \\ 2p^2 + pq + q = 5 & \text{(Equation 2)} \end{cases} \][/tex]
First, let's express [tex]\( q \)[/tex] from Equation 1:
[tex]\[ q = 1 - p \][/tex]
Now we substitute [tex]\( q \)[/tex] into Equation 2:
[tex]\[ 2p^2 + p(1 - p) + (1 - p) = 5 \][/tex]
Simplify it:
[tex]\[ 2p^2 + p - p^2 + 1 - p = 5 \][/tex]
[tex]\[ p^2 - p + 1 = 5 \][/tex]
[tex]\[ p^2 - p + 1 - 5 = 0 \][/tex]
[tex]\[ p^2 - p - 4 = 0 \][/tex]
Solve this quadratic equation:
[tex]\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \][/tex]
[tex]\[ p = \frac{1 \pm \sqrt{1 + 16}}{2} \][/tex]
[tex]\[ p = \frac{1 \pm \sqrt{17}}{2} \][/tex]
Thus, we have:
[tex]\[ p_1 = \frac{1 + \sqrt{17}}{2}, \quad p_2 = \frac{1 - \sqrt{17}}{2} \][/tex]
Now, find the corresponding [tex]\( q \)[/tex] values:
[tex]\[ q_1 = 1 - p_1 = 1 - \frac{1 + \sqrt{17}}{2} = \frac{2 - (1 + \sqrt{17})}{2} = \frac{1 - \sqrt{17}}{2} \][/tex]
[tex]\[ q_2 = 1 - p_2 = 1 - \frac{1 - \sqrt{17}}{2} = \frac{2 - (1 - \sqrt{17})}{2} = \frac{1 + \sqrt{17}}{2} \][/tex]
So, the possible pairs [tex]\((p, q)\)[/tex] are:
[tex]\[ \left( \frac{1 + \sqrt{17}}{2}, \frac{1 - \sqrt{17}}{2} \right) \quad \text{and} \quad \left( \frac{1 - \sqrt{17}}{2}, \frac{1 + \sqrt{17}}{2} \right) \][/tex]
We are given two functions: [tex]\( f(x) = 3x + 4 \)[/tex] and [tex]\( g(x) = px + q \)[/tex], where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are constants.
We need to determine the values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex] given the conditions: [tex]\( g(f(-1)) = 1 \)[/tex] and [tex]\( g^2(2) = 5 \)[/tex].
### Step 1: Calculate [tex]\( f(-1) \)[/tex]
First, we find the value of [tex]\( f(-1) \)[/tex]:
[tex]\[ f(x) = 3x + 4 \][/tex]
Substitute [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3(-1) + 4 = -3 + 4 = 1 \][/tex]
### Step 2: Calculate [tex]\( g(f(-1)) \)[/tex]
We now calculate [tex]\( g(f(-1)) \)[/tex]:
[tex]\[ f(-1) = 1 \][/tex]
Therefore:
[tex]\[ g(f(-1)) = g(1) = p(1) + q = p + q \][/tex]
Given [tex]\( g(f(-1)) = 1 \)[/tex]:
[tex]\[ p + q = 1 \quad \text{(Equation 1)} \][/tex]
### Step 3: Calculate [tex]\( g^2(2) \)[/tex]
Next, we calculate [tex]\( g^2(2) \)[/tex] which means [tex]\( g(g(2)) \)[/tex]:
[tex]\[ g(x) = px + q \][/tex]
First, find [tex]\( g(2) \)[/tex]:
[tex]\[ g(2) = p(2) + q = 2p + q \][/tex]
Then calculate [tex]\( g(g(2)) \)[/tex]:
[tex]\[ g(g(2)) = g(2p + q) = p(2p + q) + q = 2p^2 + pq + q \][/tex]
Given [tex]\( g^2(2) = 5 \)[/tex]:
[tex]\[ 2p^2 + pq + q = 5 \quad \text{(Equation 2)} \][/tex]
### Step 4: Solve the system of equations
We have the following system of equations:
[tex]\[ \begin{cases} p + q = 1 & \text{(Equation 1)} \\ 2p^2 + pq + q = 5 & \text{(Equation 2)} \end{cases} \][/tex]
First, let's express [tex]\( q \)[/tex] from Equation 1:
[tex]\[ q = 1 - p \][/tex]
Now we substitute [tex]\( q \)[/tex] into Equation 2:
[tex]\[ 2p^2 + p(1 - p) + (1 - p) = 5 \][/tex]
Simplify it:
[tex]\[ 2p^2 + p - p^2 + 1 - p = 5 \][/tex]
[tex]\[ p^2 - p + 1 = 5 \][/tex]
[tex]\[ p^2 - p + 1 - 5 = 0 \][/tex]
[tex]\[ p^2 - p - 4 = 0 \][/tex]
Solve this quadratic equation:
[tex]\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ p = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \][/tex]
[tex]\[ p = \frac{1 \pm \sqrt{1 + 16}}{2} \][/tex]
[tex]\[ p = \frac{1 \pm \sqrt{17}}{2} \][/tex]
Thus, we have:
[tex]\[ p_1 = \frac{1 + \sqrt{17}}{2}, \quad p_2 = \frac{1 - \sqrt{17}}{2} \][/tex]
Now, find the corresponding [tex]\( q \)[/tex] values:
[tex]\[ q_1 = 1 - p_1 = 1 - \frac{1 + \sqrt{17}}{2} = \frac{2 - (1 + \sqrt{17})}{2} = \frac{1 - \sqrt{17}}{2} \][/tex]
[tex]\[ q_2 = 1 - p_2 = 1 - \frac{1 - \sqrt{17}}{2} = \frac{2 - (1 - \sqrt{17})}{2} = \frac{1 + \sqrt{17}}{2} \][/tex]
So, the possible pairs [tex]\((p, q)\)[/tex] are:
[tex]\[ \left( \frac{1 + \sqrt{17}}{2}, \frac{1 - \sqrt{17}}{2} \right) \quad \text{and} \quad \left( \frac{1 - \sqrt{17}}{2}, \frac{1 + \sqrt{17}}{2} \right) \][/tex]
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