Question

How to find min and max of x|2x + 5|

Answer 1

can't imagine there is a max right?

Answer 2

The function isn't bounded above or below.

Answer 3

likewise for min. plug in values that go to minus infinity, thing goes there as well

Answer 4

IN my caclulator it says there is

Answer 5

what alchemista said (somewhat more elegantly)

Answer 6

Perhaps you are confusing local minimums with the function overall.

Answer 7

then i guess there must be. ask the calculator what it is

Answer 8

I know....For future reference though, do I use the x value or y value or both if I do get a min and max?

Answer 9

My apologies, I just came from Algebra ll and I always do hands on, calculator is a bit obtuse to me

Answer 10

Pre Calc man oh man

Answer 11

For the local minimum, you should be finding the roots of the quadratic and then between the roots will lie the minimum in this case.

Answer 12

ah i see

Answer 13

also

Answer 14

how to do you find if it is bounded above, bounded below, bonded on its domain

Answer 15

example y= 32

Answer 16

The constant function \(f(x) = 32\)?

Answer 17

y = 32 is a horizontal line. or rather the graph is. only range values isi {32}

Answer 18

but how can you tell if it is bounded, etc, or not?

Answer 19

In that case its just bounded above and below by 32

Answer 20

Its simple for all \(x\) we have \(32 \leq f(x) \leq 32\)

Answer 21

how can you tell of this one y= 2-x^2

Answer 22

what things do i look for to find if it is bounded or not

Answer 23

Well there is a maximum for that function, when x is 0

Answer 24

The maximum is 2, and the function diverges to negative infinity to the left and to the right.

Answer 25

So its bounded above by 2, but it is not bounded below.

Answer 26

ah i see. how are you doing this without technical help?

Answer 27

alright. i just got a huge impediment; y = 2^x

Answer 28

Lets talk about the previous examples first. In any case without calculus it is difficult to prove any of these things rigorously. However, lets take the previous example, which is fairly simple.

Answer 29

\(x^2 \geq 0\) for a all \(x \in \mathbb{R}\) and 0 when \(x=0\).
So we have \(f(x) = 2 - x^2\). It is clear that the function will be highest when \(x^2\) is lowest which is at 0.

Answer 30

When x is 0, the functions value is 2.

Answer 31

So the function is bounded above by 2. However, as x gets very large to the right or to the left, the function gets very becomes a very large negative number. As you approach infinity in either direction the function diverges to negative infinity.

Answer 32

seemingly irrelevant question, if infinity goes right,up or down, is it still positive?

Answer 33

so would 2^x be bounded to its domain?

Answer 34

What do you mean bounded to its domain?

Answer 35

As for your previous question, a function is said to diverge to positive infinity, if the value becomes a larger and larger \(\textbf{positive}\) number and diverges to negative infinity if the value becomes and larger and larger \(\textbf{negative}\) number.

Answer 36

sorry to deviate but my graph starts near the x-axis starting from the left curving up to infinity. fory y=2^x

Answer 37

Well first of all, clearly \(2^x\) does not have a maximum. But it also does not have a minimum. Why do you think that is?

Answer 38

It is certainly bounded below by zero. In fact zero is also the greatest lower bound. However, it does not acquire the value 0.

Answer 39

it doenst have a place on the y axis? such as 2 - 3x or maybe i need to be rectified

Answer 40

Did they talk about asymptotes in your class yet?

Answer 41

Yes

Answer 42

So it has a horizontal asymptote at 0

Answer 43

Thanks you.