Question

How do you find the domain of the function P(x) = 10^x^2 + log(1-2x)

Answer 1

since there is a logarithm in it; you have to limit the domain of "x" to values that make (1-2x)>0. Logarithms never reach 0 nor do they go negative so those values are meaningless.

1-2x>0
1>0+2x
1>2x
1>x flip it the whole thing around to read it easier and we get:

x<1

Answer 2

hit the wrong button lol

x< (1/2)

Answer 3

so the domain is 1/2?

Answer 4

the domain is {x|x<(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 (1-2x) 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.

Answer 5

Thanks! Do you do anything with 10 to the power of x^2?

Answer 6

10^(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(1-2x) are sharing the same values for "x". We have to let log(1-2x) 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?