Question

g(x) = x + 1 - e^x

find the maximum value of g(x0, and hence show that x + 1 (lesser than or equal to) e^x

Answer 1

aw hi parth :)

Answer 2

Hey. I think I know how to answer this :-)

Answer 3

ok SHOOT NOW.
so i got the max point (e^x = 1)

Answer 4

but what do i do after that?

Answer 5

Since \(e^x\) is much more rapidly increasing than \(x + 1\) is, we can already guess that the max value of \(g(x) \) is at the point \(x = 0\)

Answer 6

\[g(0) = 0 \le e^0 = 1\]

Answer 7

but i got e^x = 1. what do i do with this??

Answer 8

that is the max point.

Answer 9

ok. now what?

Answer 10

Since the maximum point, \(g(0)\), is positive, we can say that \(x + 1 \le e^x\)

Answer 11

for all \(x\)

Answer 12

sorry, i don't fully understand. Can you explain it a little more??

Answer 13

just the last part: how we got to the conclusion.

Answer 14

No, I meant to say something else. I mean that the max value is non-negative.

Answer 15

Let us assume \(f(x) = x + 1\) and \(g(x) = e^x\)

Answer 16

ok, please explain. I'm going for dinner, i'll be back in 30 mins ish, so don't worry if i don't respond. anyhoo, thank you SOO much in advance :P

Answer 17

Or rather \(f(x) = x + 1\) and \(p(x) = e^x\)

Answer 18

i REALLy need help :P kay bye.

Answer 19

Suppose that we have a number \(x_1\) where \(f(x) - g(x) \) is the maximum, and \(f(x_1) - g(x_1) > 0\) and both \(f(x),g(x)\) are increasing. Then we can say that \(g(x) \ge f(x) \) for all \(x\)

Answer 20

Can you see how this works?

Answer 21

OK, see this:

If \(g(x)\) increases as \(x\) increases, then \(x + 1\) is increasing faster than \(e^x\) and so \(x + 1 \ge e^x\).

If \(g(x)\) decreases as \(x\) increases, then \(e^x\) is increasing faster than \(x + 1\) and so \(e^x \ge x + 1\).

If \(g(x)\) stays the same all the time, then \(e^x = x + 1\). Do you see where I am getting?