# How Derivatives Affect the Graph (Proving Question)

1. Nov 5, 2009

### cjaylee

1. The problem statement, all variables and given/known data

(a) If f and g are positive, increasing, concave upward functions on I, show that the product function fg is concave upward on I.

(b)Show that part (a) remains true if f and g are both decreasing.

2. Relevant equations

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3. The attempt at a solution

(a)
f>0, f'>0, f''>0
g>0, g'>0, g''>0

(fg)'=f'g+fg'>0
(fg)''=f''g+2f'g'+fg''>0 and so fg is concave upward on I.

(b)
f'<0, f''<0
g'<0, g''<0

(fg)''=f''g+2f'g'+fg''

I have 2f'g'>0 but f''g <0 and fg''<0. Then I'm stuck here. The answer mentioned that (fg)"> or equals to f''g+fg''>0 but I ain't sure how that came about.

Thanks!

2. Nov 5, 2009

### clamtrox

They are still positive and concave, aren't they?

3. Nov 5, 2009

### cjaylee

If f and g are both decreasing, then f'<0 and g'<0. Is that right? And if both the slopes are decreasing, then f''<0 and g''<0?

4. Nov 5, 2009

### clamtrox

That does not mean that the slopes are decreasing. Suppose for example f = 1/x.

5. Nov 5, 2009

### cjaylee

It says in the text that if f'>0, then f is increasing and if f'<0, then f is decreasing on the interval?

6. Nov 5, 2009

### clamtrox

... and the text would be correct. However, f is not the slope, f is the function. f' is the slope and knowing that the slope is negative does definitely not tell you how it is changing.

7. Nov 5, 2009

### cjaylee

Oh okay. Thanks.