Determine the domain of lower h left-parenthesis x right-parenthesis equals Start Fraction 5 over Start Root 2 x plus 10 End Root End Fraction and write it using set builder notation.
To determine the domain of a function, we need to find all possible values of x for which the function is defined.
In this case, the function is defined as long as the expression under the square root sign is non-negative.
So, we set the expression greater than or equal to zero and solve for x:
2x + 10 ≥ 0
Subtract 10 from both sides:
2x ≥ -10
Divide both sides by 2:
x ≥ -5
Therefore, the domain of the function is all real numbers greater than or equal to -5.
Using set builder notation, we can write the domain as:
{x | x ≥ -5}
Find left-parenthesis f dot g right-parenthesis left-parenthesis x right-parenthesis when lower f left-parenthesis x right-parenthesis equals Start Fraction Start Root x plus 3 End Root over x End Fraction and lower g left-parenthesis x right-parenthesis equals Start Fraction Start Root x plus 3 End Root over 2 x End Fraction.
To find \(f \circ g(x)\), we need to substitute the expression for \(g(x)\) into the function \(f(x)\):
\[
f(g(x)) = f\left(\frac{{\sqrt{x + 3}}}{{2x}}\right)
\]
Now, we substitute this expression into the function \(f(x)\):
\[
f\left(\frac{{\sqrt{x + 3}}}{{2x}}\right) = \frac{{\sqrt{\frac{{\sqrt{x + 3}}}{{2x}} + 3}}}{{\frac{{\sqrt{x + 3}}}{{2x}}}}
\]
To simplify this expression further, we can simplify the numerator and denominator separately:
Numerator:
\[
\sqrt{\frac{{\sqrt{x + 3}}}{{2x}} + 3} = \sqrt{\frac{{\sqrt{x + 3} + 6x}}{{2x}}}
\]
Denominator:
\[
\frac{{\sqrt{x + 3}}}{{2x}} = \frac{{\sqrt{x + 3}}}{{2x}} \cdot \frac{{2\sqrt{x + 3}}}{{2\sqrt{x + 3}}} = \frac{{2\sqrt{x + 3} \cdot \sqrt{x + 3}}}{{2x \cdot \sqrt{x + 3}}} = \frac{{2(x + 3)}}{{2x \cdot \sqrt{x + 3}}} = \frac{{x + 3}}{{x \cdot \sqrt{x + 3}}}
\]
Now, we substitute the simplified numerator and denominator back into the expression:
\[
\frac{{\sqrt{\frac{{\sqrt{x + 3} + 6x}}{{2x}} + 3}}}{{\frac{{\sqrt{x + 3}}}{{2x}}}} = \frac{{\sqrt{\frac{{\sqrt{x + 3} + 6x}}{{2x}} + 3}}}{{\frac{{x + 3}}{{x \cdot \sqrt{x + 3}}}}}
\]
Thus, \(f \circ g (x) = \frac{{\sqrt{\frac{{\sqrt{x + 3} + 6x}}{{2x}} + 3}}}{{\frac{{x + 3}}{{x \cdot \sqrt{x + 3}}}}}\).
choices
a) left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction x squared plus 6 x plus 9 over 2 x squared End FractionImage with alt text: left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction x squared plus 6 x plus 9 over 2 x squared End Fraction
b) left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction Start Root x plus 3 End Root over 2 x squared End Fraction
Image with alt text: left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction Start Root x plus 3 End Root over 2 x squared End Fraction
c) left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction x plus 3 over 2 x squared End Fraction
Image with alt text: left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction x plus 3 over 2 x squared End Fraction
d) left-parenthesis lower f dot lower g right-parenthesis left-parenthesis x right-parenthesis equals Start Fraction x squared plus 9 over 2 x squared End Fraction
The correct choice is:
b) \(f \circ g (x) = \frac{{\sqrt{x + 3}}}{{2x^2}}\)
Determine the domain of extended-left-parenthesis Start Fraction f over g End Fraction extended-right-parenthesis left-parenthesis x right-parenthesis when lower f left-parenthesis x right-parenthesis equals negative Start Fraction 1 over x End Fraction and lower g left-parenthesis x right-parenthesis equals Start Root 3 x minus 9 End Root
To determine the domain of \( \frac{f}{g}(x) \), we need to consider two factors:
1) The denominator cannot equal zero since division by zero is undefined.
2) The value of \( \frac{f}{g}(x) \) must be a real number.
First, let's find the values of x that make the denominator, \( g(x) \), equal to zero:
\[ \sqrt{3x - 9} = 0 \]
To solve this equation, we isolate x by squaring both sides and solving for x:
\[ 3x - 9 = 0 \]
\[ 3x = 9 \]
\[ x = 3 \]
Therefore, x = 3 is the value that makes the denominator zero. Therefore, we need to exclude x = 3 from the domain.
Next, we need to consider the values of x that would make \( \frac{f}{g}(x) \) a real number. Since both the numerator and denominator of \( \frac{f}{g}(x) \) involve square roots, the only concern is whether the values inside the square roots are non-negative.
For the numerator, \( f(x) = -\frac{1}{x} \), there are no restrictions, as long as x is a non-zero real number.
For the denominator, \( g(x) = \sqrt{3x - 9} \), we need the expression inside the square root to be non-negative:
\[ 3x - 9 \geq 0 \]
\[ x \geq 3 \]
Combining both conditions, we need x to be a non-zero real number and x to be greater than 3 (excluding x = 3).
Therefore, the domain of \( \frac{f}{g}(x) \) is: \( x \in (-\infty, 0) \cup (3, \infty) \).
choices
a) left-parenthesis 3 comma infinity right-parenthesisImage with alt text: left-parenthesis 3 comma infinity right-parenthesis
b) left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis 0 comma 3 right-parenthesis union left-parenthesis 3 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis 0 comma 3 right-parenthesis union left-parenthesis 3 comma infinity right-parenthesis
c) left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis 3 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma 0 right-parenthesis union left-parenthesis 3 comma infinity right-parenthesis
d) left-parenthesis negative infinity comma 3 right-parenthesis union left-parenthesis 3 comma infinity right-parenthesis
The correct choice is:
c) \(\left(-\infty, 0\right) \cup \left(3, \infty\right)\)
To determine the domain of a function, we need to find the set of all possible values that x can take. In this case, we have the function h(x) = 5 / √(2x + 10).
To find the domain, we need to consider any restrictions on the function. In this case, we have the square root (√) in the denominator. The square root of a negative number is not defined in the real number system. Therefore, the expression 2x + 10 inside the square root must be non-negative.
To find the values that make 2x + 10 non-negative, we solve the inequality:
2x + 10 ≥ 0
Subtracting 10 from both sides, we get:
2x ≥ -10
Dividing both sides by 2, we have:
x ≥ -5
So, the expression 2x + 10 must be greater than or equal to zero, which means x must be greater than or equal to -5.
Therefore, the domain of the function h(x) is the set of all real numbers x such that x ≥ -5.
In set-builder notation, this domain can be written as:
{ x : x ≥ -5 }
Here, the condition x ≥ -5 represents the set of all x-values that satisfy the inequality.