# Homework Help: More logs help =(

1. Jul 14, 2005

### bengalibabu

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

2. Jul 15, 2005

### toocool_sashi

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???

3. Jul 15, 2005

### trinitron

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.

4. Jul 15, 2005

### Pyrrhus

For the first question, you can obtain m (the slope) at (1,0) throught 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.

5. Jul 15, 2005

### BobG

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:

$$e^x+xe^x=0$$
$$e^x=-xe^x$$

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