Answer :
To find the [tex]\(y\)[/tex]-intercept of the graph of the function [tex]\( f(x) = \frac{20}{1 + 9 e^{3x}} \)[/tex], we need to determine the value of the function when [tex]\( x = 0 \)[/tex]. The [tex]\(y\)[/tex]-intercept is the point where the graph intersects the [tex]\( y \)[/tex]-axis, which happens when [tex]\( x = 0 \)[/tex].
1. Start by substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{20}{1 + 9 e^{3 \cdot 0}} \][/tex]
2. Simplify the exponent in the denominator:
[tex]\[ e^{3 \cdot 0} = e^0 = 1 \][/tex]
3. Substitute [tex]\( e^0 = 1 \)[/tex] back into the equation:
[tex]\[ f(0) = \frac{20}{1 + 9 \cdot 1} \][/tex]
4. Simplify the expression within the denominator:
[tex]\[ 1 + 9 \cdot 1 = 1 + 9 = 10 \][/tex]
5. Perform the division:
[tex]\[ f(0) = \frac{20}{10} = 2 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept of the graph of [tex]\( f(x) \)[/tex] is [tex]\( 2 \)[/tex].
[tex]\[ \boxed{2} \][/tex]
1. Start by substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{20}{1 + 9 e^{3 \cdot 0}} \][/tex]
2. Simplify the exponent in the denominator:
[tex]\[ e^{3 \cdot 0} = e^0 = 1 \][/tex]
3. Substitute [tex]\( e^0 = 1 \)[/tex] back into the equation:
[tex]\[ f(0) = \frac{20}{1 + 9 \cdot 1} \][/tex]
4. Simplify the expression within the denominator:
[tex]\[ 1 + 9 \cdot 1 = 1 + 9 = 10 \][/tex]
5. Perform the division:
[tex]\[ f(0) = \frac{20}{10} = 2 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept of the graph of [tex]\( f(x) \)[/tex] is [tex]\( 2 \)[/tex].
[tex]\[ \boxed{2} \][/tex]
