$y=x \cdot e^{\frac{x}{10}}$
Answer (2)
Find the first derivative using the product rule: d x d y = e 10 x ( 1 + 10 x ) .
Find the second derivative: d x 2 d 2 y = 10 1 e 10 x ( 2 + 10 x ) .
Determine critical points by solving d x d y = 0 , which gives x = − 10 .
Determine inflection points by solving d x 2 d 2 y = 0 , which gives x = − 20 .
The function's properties are then analyzed based on these points.
The \textbf{critical point} is x = − 10 and the \textbf{inflection point} is x = − 20 .
The \textbf{x and y intercepts} are both 0.
The \textbf{limit} as x approaches infinity is infinity, and as x approaches negative infinity is 0.
The final answer is the analysis of the function's properties. See analysis in steps
Explanation
Problem Analysis We are given the function y = x \tento e 10 x and asked to analyze its properties. This involves finding its first and second derivatives, critical points, inflection points, intervals of increasing and decreasing behavior, concavity, and limits as x approaches infinity.
Finding the First Derivative First, we find the first derivative of y with respect to x using the product rule: d x d y = d x d ( x e 10 x ) = e 10 x + x ⋅ 10 1 e 10 x = e 10 x ( 1 + 10 x )
Finding the Second Derivative Next, we find the second derivative of y with respect to x :
d x 2 d 2 y = d x d ( e 10 x ( 1 + 10 x ) ) = 10 1 e 10 x ( 1 + 10 x ) + e 10 x ⋅ 10 1 = 10 1 e 10 x ( 2 + 10 x )
Finding Critical Points To find the critical points, we set d x d y = 0 :
e 10 x ( 1 + 10 x ) = 0 Since e 10 x is never zero, we have: 1 + 10 x = 0 ⟹ x = − 10 Thus, the critical point is x = − 10 .
Increasing and Decreasing Intervals To determine the intervals where the function is increasing or decreasing, we analyze the sign of d x d y :
If x < − 10 , then 1 + 10 x < 0 , so d x d y < 0 , and the function is decreasing.
If -10"> x > − 10 , then 0"> 1 + 10 x > 0 , so 0"> d x d y > 0 , and the function is increasing.
Finding Inflection Points To find the inflection points, we set d x 2 d 2 y = 0 :
10 1 e 10 x ( 2 + 10 x ) = 0 Since e 10 x is never zero, we have: 2 + 10 x = 0 ⟹ x = − 20 Thus, the inflection point is x = − 20 .
Determining Concavity To determine the concavity of the function, we analyze the sign of d x 2 d 2 y :
If x < − 20 , then 2 + 10 x < 0 , so d x 2 d 2 y < 0 , and the function is concave down.
If -20"> x > − 20 , then 0"> 2 + 10 x > 0 , so 0"> d x 2 d 2 y > 0 , and the function is concave up.
Limits at Infinity Now, let's analyze the behavior of the function as x approaches positive and negative infinity: x → ∞ lim x e 10 x = ∞ x → − ∞ lim x e 10 x = 0
Finding Intercepts Finally, let's find the x and y intercepts:
x -intercept: y = 0 ⟹ x e 10 x = 0 ⟹ x = 0
y -intercept: x = 0 ⟹ y = 0 e 10 0 = 0 Thus, the x and y intercepts are both at x = 0 and y = 0 .
Summary of Analysis In summary:
Critical point: x = − 10 , y = − 10 e − 1 ≈ − 3.679
Inflection point: x = − 20 , y = − 20 e − 2 ≈ − 2.707
Decreasing for x < − 10 , increasing for -10"> x > − 10
Concave down for x < − 20 , concave up for -20"> x > − 20
lim x → ∞ x e 10 x = ∞
lim x → − ∞ x e 10 x = 0
x -intercept: x = 0
y -intercept: y = 0
Examples
Understanding the behavior of functions like y = x \tento e 10 x is crucial in various fields. For instance, in physics, this type of function can model damped oscillations or the decay of radioactive substances. In economics, it can represent growth models where the rate of growth decreases over time. By analyzing critical points, inflection points, and limits, we can predict and optimize system behaviors, leading to more efficient designs and strategies.
The function y = x ⋅ e 10 x has a critical point at x = − 10 and an inflection point at x = − 20 . It is decreasing for x < − 10 and increasing for -10"> x > − 10 , with concavity changing at x = − 20 . Both intercepts occur at the origin ( 0 , 0 ) .
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