Solve Questions on y=x^2 ln x and f(x) = xe^x

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Homework Help Overview

The discussion revolves around two calculus-related questions: finding the equation of the tangent line to the curve y = x^2 ln x at a specific point and determining critical points for the function f(x) = xe^x, including its minimum value and points of inflection.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to differentiate the function to find the slope for the tangent line and evaluate it at the given point. There is also mention of using the product rule for differentiation. For the second question, the concept of critical points is raised, with suggestions to set the derivative to zero and analyze the results for maxima, minima, and points of inflection.

Discussion Status

Several participants have provided hints and guidance on how to approach the problems, emphasizing the importance of derivatives and critical points. There is an ongoing exploration of methods without reaching a consensus on specific solutions.

Contextual Notes

Participants express uncertainty about their understanding of calculus concepts, particularly regarding derivatives and their implications for the problems at hand.

bengalibabu
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hey guys...i just had 2 questions that i really needed help with, i have no clue how to even start these question, any help would be greatly appreciated, thx

Question 1
Find the equation of the tangent line to the curve y = x^2 ln x at the point (1,0).
the answer should be in the form y = mx + b

Question 2
For the functions f(x) = xe^x
1. The minimum value of the function occurs when x = _____.
2. A point of inflection occurs when x = ______.


thx again

bengalibabu
 
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hmmm did u do differential calculus?? ill give u some broad hints...firstly both questions are not log based they are calculus based.

Question 1
To uniquely determine any line, we need 2 things, slope and a point thru which it passes. the second condition is already given to us (1,0). To get slope, think abt the geometric implication of the derivative of a function?

ill tell u the hint to question 2 when u tell me ... have u done maxima and minima in calculus yet?
 
For 1 you differentiate the given function and evaluate the result at (1,0). This will give you the slope of the line, and since you have the point (x,f(x)) you now have a point and a slope and so you can find the line.

For 2 set the derivative equal to zero to find critical points. Then plug in values around each critical point to tell whether it is a minimum, maximum or point of inflection.
 
For the first question, you can obtain m (the slope) at (1,0) through the first derivative of the given function. After you find the slope with the form y = mx +b, you can find b.

For the second question, find the critical points (possible max and min), and check them with the 2nd derivative for max or min. Solve for x in the 2nd derivative for the point of inflexion.
 
bengalibabu said:
hey guys...i just had 2 questions that i really needed help with, i have no clue how to even start these question, any help would be greatly appreciated, thx

Question 1
Find the equation of the tangent line to the curve y = x^2 ln x at the point (1,0).
the answer should be in the form y = mx + b

Question 2
For the functions f(x) = xe^x
1. The minimum value of the function occurs when x = _____.
2. A point of inflection occurs when x = ______.


thx again

bengalibabu
The others pretty much covered it. For 1, find the derivative and evaluate for x=1. You need the product rule to find the derivative. x=1 turns out to be a very easy place to evalutate the derivative at.

At the second, you set the derivative to zero and solve for x. A little tougher, but, if you think about it, there's obviously only one value for x that would give you zero for an answer. Even if you don't notice right off the bat, you can move one term to the other side:

[tex]e^x+xe^x=0[/tex]
[tex]e^x=-xe^x[/tex]

Cancel out like terms and the answer gets a whole lot more obvious.
 

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