h(x) = {2x+1 (x-1)} Find h(-2) How would I do this? 2(-2)+1 = -3 which is less than or equal to -1, but (-2)^2+2=6 which is more than -1. Does this mean it's an empty set?

Question

h(x) = {2x+1 (x<=-1), x^2+2 (x>-1)}
Find h(-2)
How would I do this? 2(-2)+1 = -3 which is less than or equal to -1, but (-2)^2+2=6 which is more than -1. Does this mean it's an empty set?

anonymous · Answer

@lgbasallote

@jim_thompson5910

anonymous · Answer

@campbell_st
@phi

@estudier

@radar

@Chlorophyll

@AccessDenied

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@zzr0ck3r

@eigenschmeigen

anonymous · Answer

-_-

anonymous · Answer

Please? :)

campbell_st · Answer

you are correct with your assumption

h(-2) = 2(-2) + 1

anonymous · Answer

ok, team effort guys! we're all tagged so each of us has to contribute one word!

anonymous · Answer

my word is 

apple

anonymous · Answer

Which one is right? They are both right

campbell_st · Answer

you only look at the part where the inequality holds true... and for x = -2

it is true when h(x) = 2x + 1

anonymous · Answer

but it's also true when  x^2+2

anonymous · Answer

I mean x>-1

campbell_st · Answer

no its not the 2nd part

x^2 + 2 needs an x value > -1  

and -2 is not greater than -1 

so its only the 1st part

anonymous · Answer

Wait what? I thought the output had to satisfy the inequality, not the input?

campbell_st · Answer

the output is h(x) the input is x.... so apply the condition for x...

anonymous · Answer

Alright. Wow i've been doing this wrong all this time

campbell_st · Answer

x is called the independetn variable... h(x) or y is the dependent

anonymous · Answer

Wait... then what is h(0)?

campbell_st · Answer

hope it helps

H(0) so x = 0.... use 

x^2 + 2 since  0 > -1

anonymous · Answer

OOOH. Thank you.