Solutions of this inequality

[itsrayex]

Got this inequality:

The question is "the solutions of this inequality are:
A. a closed and limited interval
B. an open unlimited interval
C. two closed unlimited intervals
D. an open limited interval"

Honestly I have no clue how to go about this. I know that e^x is always positive and -x^2 is always negative but idk how that can help me answering the question... can someone please help?

 

[Otis]

-x^2 is always negative

That's true, itsrayex. However, you've been given 2-x^2, which means the graph of -x^2 has been shifted vertically upwards two units. Draw a rough sketch of both y=e^x and y=2-x^2 together, and see what you think. (If you're not familiar with the shapes of these basic curves, check your textbook.)



[imath]\;[/imath]

 

[Deleted member 4993]

Got this inequality:
View attachment 31083
The question is "the solutions of this inequality are:
A. a closed and limited interval
B. an open unlimited interval
C. two closed unlimited intervals
D. an open limited interval"

Honestly I have no clue how to go about this. I know that e^x is always positive and -x^2 is always negative but idk how that can help me answering the question... can someone please help?

As suggested in response #2, plotting y = e^x +x² -2 is an excellent step to generate ideas to solve the in equality. To save time yo may want to use wolframalpha.com for a quick plot or your graphing calculator.

 

[itsrayex]

That's true, itsrayex. However, you've been given 2-x^2, which means the graph of -x^2 has been shifted vertically upwards two units. Draw a rough sketch of both y=e^x and y=2-x^2 together, and see what you think. (If you're not familiar with the shapes of these basic curves, check your textbook.)



[imath]\;[/imath]

Thank you for your answer.
So, since the function has been shifted vertically it means a little part of it is on the positive part of the plane while the rest is on the negative part, right? While as already said e^x is always positive. Does this mean that e^x can be smaller than 2-x^2 only in the positive part of 2-x^2? Then, the correct answer is A, a closed limited interval. Is this correct?

 

[lex]

You want to know when [imath]e^x+x^2≤2[/imath]
Clearly [imath]e^x+x^2[/imath] tends to infinity as x becomes large and positive and as x becomes large and negative, so the solution to [imath]e^x+x^2≤2[/imath] doesn't contain an unlimited interval. This leaves only A or D.
The solutions to [imath]e^x+x^2≤2[/imath] will contain endpoints, because of the equality, therefore it must be A.

 

[lex]

Thank you for your answer.
So, since the function has been shifted vertically it means a little part of it is on the positive part of the plane while the rest is on the negative part, right? While as already said e^x is always positive. Does this mean that e^x can be smaller than 2-x^2 only in the positive part of 2-x^2? Then, the correct answer is A, a closed limited interval. Is this correct?

Your logic is correct.

 

[Otis]

the positive part of the plane

Be careful using the phrase "positive part" to describe a region in the xy-plane. I understand that you're thinking only of y-coordinates, but some people might not.

Your answer is correct. The solution interval is closed because both endpoints are included. The solution is limited because it does not include values that are infinite in absolute value.



[imath]\;[/imath]