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anonymous
 5 years ago
How do you find the domain of the function P(x) = 10^x^2 + log(12x)
anonymous
 5 years ago
How do you find the domain of the function P(x) = 10^x^2 + log(12x)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since there is a logarithm in it; you have to limit the domain of "x" to values that make (12x)>0. Logarithms never reach 0 nor do they go negative so those values are meaningless. 12x>0 1>0+2x 1>2x 1>x flip it the whole thing around to read it easier and we get: x<1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hit the wrong button lol x< (1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the domain is 1/2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the domain is {xx<(1/2)} in other words. x can be any value that is less than (1/2) another notation would be (inf,(1/2)) think about it this way. In order for (12x) to be 0 (bad number; BAD number) "x" would have to be (1/2). Anything bigger than this would make (2)(+number) = a negative number greater than 1. For instance: 1  2(1) = 1. Cant have negatives in the log function!!! log(1) is strictly forbidden and is punishable by.... by.... well, something horrible I am sure. So, anything equal to and over (1/2) has got to be removed from the domain. That leaves everything less than (1/2) to negative infinity that CAN be used.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks! Do you do anything with 10 to the power of x^2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.010^(x^2) has no limitations. In other words, the input value of an exponential function can be any "real" number and it will output a valid answer. But, because 10^(x^2) and log(12x) are sharing the same values for "x". We have to let log(12x) control what we can use as inputs. A domain tells us what values we can use for inputs in a function. Domain = Inputs. Make sense?
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