Chapter3: Functions
Section3.5: Transformation Of Functions
Problem 5SE: How can you determine whether a function is odd or even from the formula of the function?
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![Title: Finding the Tangent Line to a Curve at a Given Point
Problem Statement:
Find the equation of the tangent line to the curve \( y = \frac{3 + x}{4 + e^x} \) where \( x = 0 \).
To solve this problem, we need to follow these steps:
1. **Find the value of the function at \( x = 0 \).**
Substitute \( x = 0 \) into the given equation:
\[
y = \frac{3 + 0}{4 + e^0} = \frac{3}{4 + 1} = \frac{3}{5}
\]
So, the point of tangency is \((0, \frac{3}{5})\).
2. **Find the derivative of the function to determine the slope of the tangent line.**
The given function is \( y = \frac{3 + x}{4 + e^x} \). To find the derivative \( y' \), use the quotient rule:
\[
y' = \frac{(u'v - uv')}{v^2}
\]
where \( u = 3 + x \) and \( v = 4 + e^x \).
First, compute the derivatives \( u' \) and \( v' \):
\[
u' = 1
\]
\[
v' = e^x
\]
Now, apply the quotient rule:
\[
y' = \frac{(1)(4 + e^x) - (3 + x)(e^x)}{(4 + e^x)^2}
\]
Simplify the numerator:
\[
y' = \frac{4 + e^x - 3e^x - xe^x}{(4 + e^x)^2} = \frac{4 + e^x - 3e^x - xe^x}{(4 + e^x)^2} = \frac{4 - 2e^x - xe^x}{(4 + e^x)^2}
\]
3. **Evaluate the derivative at \( x = 0 \).**
Substitute \( x = 0 \) into the derivative:
\[
y'(0) = \frac{4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff65fb7b2-ba85-4e67-8fe1-01f97a564d1e%2F3739fb47-cb33-4122-ad8f-21518e4d1d23%2Fn48979h_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Finding the Tangent Line to a Curve at a Given Point
Problem Statement:
Find the equation of the tangent line to the curve \( y = \frac{3 + x}{4 + e^x} \) where \( x = 0 \).
To solve this problem, we need to follow these steps:
1. **Find the value of the function at \( x = 0 \).**
Substitute \( x = 0 \) into the given equation:
\[
y = \frac{3 + 0}{4 + e^0} = \frac{3}{4 + 1} = \frac{3}{5}
\]
So, the point of tangency is \((0, \frac{3}{5})\).
2. **Find the derivative of the function to determine the slope of the tangent line.**
The given function is \( y = \frac{3 + x}{4 + e^x} \). To find the derivative \( y' \), use the quotient rule:
\[
y' = \frac{(u'v - uv')}{v^2}
\]
where \( u = 3 + x \) and \( v = 4 + e^x \).
First, compute the derivatives \( u' \) and \( v' \):
\[
u' = 1
\]
\[
v' = e^x
\]
Now, apply the quotient rule:
\[
y' = \frac{(1)(4 + e^x) - (3 + x)(e^x)}{(4 + e^x)^2}
\]
Simplify the numerator:
\[
y' = \frac{4 + e^x - 3e^x - xe^x}{(4 + e^x)^2} = \frac{4 + e^x - 3e^x - xe^x}{(4 + e^x)^2} = \frac{4 - 2e^x - xe^x}{(4 + e^x)^2}
\]
3. **Evaluate the derivative at \( x = 0 \).**
Substitute \( x = 0 \) into the derivative:
\[
y'(0) = \frac{4
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