Calculus Homework Help?

Question

I need help wiht this problem: Suppose g is a differntiable function that satisfies the following three properties: g is concave down
g(1)=11
g(7)=5

(a) What is the average rate of change of g on the interval 1 less than or equal to x is less than or equal to 7?
(b) Which is larger, g'(3) or g'(5)? Explain?

Answer 1

The 1'st one is easy. The average ratio of change is
(g(7) - g(1))/(7 - 1) = (5 - 11)/(7 - 1) = -1

Since g(x) is concave downwards, the slope is constantly decreasing, so g'(3) > g'(5).

HTH ☺


Doug

Answer 2

Hi.

For (a), wouldn't that be just [g(7) - g(1)]/(7-1), or basically -1?

For (b), g'(3) would be higher than g'(5). To see, we need to consider several cases. But before that, we need to have a general idea of g(x). Note that g(x) is concave down, and that g(1) is higher than g(7). This means that the absolute maximum value would lie somewhere in between x =1 and x =7, or at g(1) itself.

Case 1: g(3) occurs before the maximum and g(5) may be the maximum point, or if not, it lies before reaching maximum point

If g(5) is the maximum point, g'(5) is equal to zero. Thus, any point (which is near to it) at its right would be part of the increasing part of the function. Thus, if g(3) is part of the increasing part, then g'(3) >0. Remember that the first derivative is positive if the function is increasing. If g(5) is not yet the maximum value, then g'(5) is also >0. But because g(5) is closer to the maximum value than g(3), then g'(5) would be a positive number closer to zero than g'(3). Thus, g'(3) > g'(5).

Case 2: g(3) lies before the maximum, or it may be the maximum, and g(5) lies after the maximum value

Since g(5) lies after the maximum value, then it will be part of the decreasing part of the function, whose first derivative is <0. Thus g'(5)<0. g(3), on the other hand will be either the maximum value, or before it, thus, g'(3) = 0, or g'(3) > 0, respectively. Any way g(3) is placed here, g'(3) > g'(5).

Case 3: Both g(3) and g(5) lie after the maximum value

If both g(3) and g(5) lie after the maximum value, then both g'(3) and g'(5) <0 (being part of the decreasing part of the function). But note that, g(3) is closer to the maximum than g(5), so g'(3) would be closer to zero than g'(5). Because the "more negative" a number is the less value it has (e.g -2 is more negative than -1, thus, -2<-1), g'(3)>g'(5).

So, any case we consider, g'(3)>g'(5).

Answer 3

(a) this one is kinda easy. I think all u have to do is find the slope between those two points. (1,11) and (7,5). The slope you'll get will be (11-5)/(7-1)= 6/6. so that's your average rate of change.
(b) we know that g is concave down. so the second derivative will surely be negative. Integrate second derivative, and you'll get first derivative. and our first derivative, we can use the y=x(since our slope was one). Wait, since at point 1, it's 11 and at point 7 it's 5, and they are concaved down, we can conclude that at point g'3 is bigger than g'5. I don't really know how to do b portion, but I hope that helps.

Answer 4

(a) average rate of change = slope of a line going through the two points.

(b) concave down → decreasing derivative.

Good luck!