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you're in calc too? lol
right i need help
i suck at calc lols i think u set both -x-3 and x-5 to zero and solve
with calc you can find min/max using criticle point
how do i do that
to find criticel point , take derivative and set what you get to zero
can u write it out..
he means critical points
y=-(x+3)(x-5) let's use product rule to differentiate - (x+3)'(x-5)+(x-3)(x+5)'= - 1(x-5)+(x-3)
king explains it well here
idont udnerstand kings way. haha
i understood it :(
can u guys help me with wat i attached above i need help on number 6a-f
what do you think is domain for 6) a) in other word is there any number that you can't put in for x?
anything above 6 ?
why not anything below 6 into f(x)=5+3x
i dont know
could it be anything
you can put anything number and you won't get in problem so your domain is all real number
wwhat about range
domain is number you put in for x , range is number you get back as y as result of you putting value for x
if you can put all value for x, you will get any value of y so range is also all real number
ok now b
b is same thing
so is c
d) is different tell me what number can't we put in?
what do you mean?
in the calculator ..
for what number did you get an error?
-1 -2 -3 -4 and all negative
ok , so we can't put negative number in there , 0 is fine so domain is x must be greater or equal to 0 x>=0
how bout range
square root only return positive values sqrt=0 sqrt=1 sqrt=2 ect....... so range is too y>=0
alright, I forgot to do range for b and c
y= x^2 y of that function goes from 0 to infinity if we shift that up y=x^2+4 y goes from 4 to infinity
for c) well just look the graph http://www.wolframalpha.com/input/?i=(1.5)%5Ex&t=crmtb01 goes from 0 to infinity
i need e nd f
what do you think?
any number for domain
y=-x^2 goes from -negative infinity to 0 so y=-x^2+2 is shifted up by two
how about f
what numbers doesn't works?
any number below o
no, only number that doesn't work is 0
how about range
all real number
i need help explainging what 5 means a-d
make another thread