Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Sure, let's tackle each part of the question step by step.
### Part (a): Domain of \( g(x) = \frac{2x - 1}{5x + 3} \)
The domain of a function consists of all the possible input values (x-values) for which the function is defined. For the function \( g(x) = \frac{2x - 1}{5x + 3} \), the denominator \( 5x + 3 \) must not be zero because division by zero is undefined.
1. Set the denominator equal to zero and solve for \( x \):
[tex]\[ 5x + 3 = 0 \][/tex]
[tex]\[ 5x = -3 \][/tex]
[tex]\[ x = -\frac{3}{5} \][/tex]
2. Therefore, \( g(x) \) is undefined at \( x = -\frac{3}{5} \) and the domain of \( g(x) \) is all real numbers except \( x = -\frac{3}{5} \).
In interval notation, the domain of \( g(x) \) is:
[tex]\[ (-\infty, -\frac{3}{5}) \cup (-\frac{3}{5}, \infty) \][/tex]
### Part (b): Determine if \( g \) is onto
To determine if \( g(x) = \frac{2x - 1}{5x + 3} \) is onto, we need to check if every possible output value (y-value) has a corresponding input value (x-value) such that \( y = g(x) \). Specifically, we need to determine if the function covers all real numbers as outputs.
1. Set up the equation \( y = \frac{2x - 1}{5x + 3} \) and solve for \( x \):
[tex]\[ y(5x + 3) = 2x - 1 \][/tex]
[tex]\[ 5xy + 3y = 2x - 1 \][/tex]
[tex]\[ 5xy - 2x = -3y - 1 \][/tex]
[tex]\[ x(5y - 2) = -3y - 1 \][/tex]
[tex]\[ x = \frac{-3y - 1}{5y - 2} \][/tex]
2. To ensure that there exists an \( x \) for every \( y \), we need the expression \( \frac{-3y - 1}{5y - 2} \) to be defined for all real numbers \( y \). This expression is defined for all \( y \) except where the denominator \( 5y - 2 \) is zero:
[tex]\[ 5y - 2 = 0 \][/tex]
[tex]\[ y = \frac{2}{5} \][/tex]
Thus, the function is not onto because the function cannot produce the value \( y = \frac{2}{5} \). Hence, \( g(x) \) is not onto.
### Part (c): Range of \( x \) for which \( 2x^2 + x - 6 < 0 \)
To find the range of \( x \) for which \( 2x^2 + x - 6 < 0 \), we need to solve the inequality \( 2x^2 + x - 6 < 0 \).
1. First, solve the quadratic equation \( 2x^2 + x - 6 = 0 \) to find its roots:
[tex]\[ 2x^2 + x - 6 = 0 \][/tex]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -6 \):
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{4} \][/tex]
[tex]\[ x = \frac{-1 \pm 7}{4} \][/tex]
[tex]\[ x_1 = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2} \][/tex]
[tex]\[ x_2 = \frac{-1 - 7}{4} = \frac{-8}{4} = -2 \][/tex]
2. The roots \( x = \frac{3}{2} \) and \( x = -2 \) divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, \frac{3}{2}) \), and \( (\frac{3}{2}, \infty) \).
3. Test a value from each interval to determine where the quadratic expression is negative:
- For \( x \in (-\infty, -2) \): Test \( x = -3 \)
[tex]\[ 2(-3)^2 + (-3) - 6 = 18 - 3 - 6 = 9 > 0 \][/tex]
- For \( x \in (-2, \frac{3}{2}) \): Test \( x = 0 \)
[tex]\[ 2(0)^2 + 0 - 6 = -6 < 0 \][/tex]
- For \( x \in (\frac{3}{2}, \infty) \): Test \( x = 2 \)
[tex]\[ 2(2)^2 + 2 - 6 = 8 + 2 - 6 = 4 > 0 \][/tex]
4. Hence, the quadratic expression \( 2x^2 + x - 6 \) is negative in the interval \( (-2, \frac{3}{2}) \).
In interval notation, the solution is:
[tex]\[ x \in \left( -2, \frac{3}{2} \right) \][/tex]
To summarize:
- The domain of \( g(x) \) is \( (-\infty, -\frac{3}{5}) \cup (-\frac{3}{5}, \infty) \).
- The function \( g(x) \) is not onto.
- The range of [tex]\( x \)[/tex] for which [tex]\( 2x^2 + x - 6 < 0 \)[/tex] is [tex]\( \left( -2, \frac{3}{2} \right) \)[/tex].
### Part (a): Domain of \( g(x) = \frac{2x - 1}{5x + 3} \)
The domain of a function consists of all the possible input values (x-values) for which the function is defined. For the function \( g(x) = \frac{2x - 1}{5x + 3} \), the denominator \( 5x + 3 \) must not be zero because division by zero is undefined.
1. Set the denominator equal to zero and solve for \( x \):
[tex]\[ 5x + 3 = 0 \][/tex]
[tex]\[ 5x = -3 \][/tex]
[tex]\[ x = -\frac{3}{5} \][/tex]
2. Therefore, \( g(x) \) is undefined at \( x = -\frac{3}{5} \) and the domain of \( g(x) \) is all real numbers except \( x = -\frac{3}{5} \).
In interval notation, the domain of \( g(x) \) is:
[tex]\[ (-\infty, -\frac{3}{5}) \cup (-\frac{3}{5}, \infty) \][/tex]
### Part (b): Determine if \( g \) is onto
To determine if \( g(x) = \frac{2x - 1}{5x + 3} \) is onto, we need to check if every possible output value (y-value) has a corresponding input value (x-value) such that \( y = g(x) \). Specifically, we need to determine if the function covers all real numbers as outputs.
1. Set up the equation \( y = \frac{2x - 1}{5x + 3} \) and solve for \( x \):
[tex]\[ y(5x + 3) = 2x - 1 \][/tex]
[tex]\[ 5xy + 3y = 2x - 1 \][/tex]
[tex]\[ 5xy - 2x = -3y - 1 \][/tex]
[tex]\[ x(5y - 2) = -3y - 1 \][/tex]
[tex]\[ x = \frac{-3y - 1}{5y - 2} \][/tex]
2. To ensure that there exists an \( x \) for every \( y \), we need the expression \( \frac{-3y - 1}{5y - 2} \) to be defined for all real numbers \( y \). This expression is defined for all \( y \) except where the denominator \( 5y - 2 \) is zero:
[tex]\[ 5y - 2 = 0 \][/tex]
[tex]\[ y = \frac{2}{5} \][/tex]
Thus, the function is not onto because the function cannot produce the value \( y = \frac{2}{5} \). Hence, \( g(x) \) is not onto.
### Part (c): Range of \( x \) for which \( 2x^2 + x - 6 < 0 \)
To find the range of \( x \) for which \( 2x^2 + x - 6 < 0 \), we need to solve the inequality \( 2x^2 + x - 6 < 0 \).
1. First, solve the quadratic equation \( 2x^2 + x - 6 = 0 \) to find its roots:
[tex]\[ 2x^2 + x - 6 = 0 \][/tex]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -6 \):
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{4} \][/tex]
[tex]\[ x = \frac{-1 \pm 7}{4} \][/tex]
[tex]\[ x_1 = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2} \][/tex]
[tex]\[ x_2 = \frac{-1 - 7}{4} = \frac{-8}{4} = -2 \][/tex]
2. The roots \( x = \frac{3}{2} \) and \( x = -2 \) divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, \frac{3}{2}) \), and \( (\frac{3}{2}, \infty) \).
3. Test a value from each interval to determine where the quadratic expression is negative:
- For \( x \in (-\infty, -2) \): Test \( x = -3 \)
[tex]\[ 2(-3)^2 + (-3) - 6 = 18 - 3 - 6 = 9 > 0 \][/tex]
- For \( x \in (-2, \frac{3}{2}) \): Test \( x = 0 \)
[tex]\[ 2(0)^2 + 0 - 6 = -6 < 0 \][/tex]
- For \( x \in (\frac{3}{2}, \infty) \): Test \( x = 2 \)
[tex]\[ 2(2)^2 + 2 - 6 = 8 + 2 - 6 = 4 > 0 \][/tex]
4. Hence, the quadratic expression \( 2x^2 + x - 6 \) is negative in the interval \( (-2, \frac{3}{2}) \).
In interval notation, the solution is:
[tex]\[ x \in \left( -2, \frac{3}{2} \right) \][/tex]
To summarize:
- The domain of \( g(x) \) is \( (-\infty, -\frac{3}{5}) \cup (-\frac{3}{5}, \infty) \).
- The function \( g(x) \) is not onto.
- The range of [tex]\( x \)[/tex] for which [tex]\( 2x^2 + x - 6 < 0 \)[/tex] is [tex]\( \left( -2, \frac{3}{2} \right) \)[/tex].
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
