>>12994136>e^(1-x^2) draws a positive line through all values.2 > 0
2 * 2 = 2^2 > 0
2 * 2 * 2 = 2^3 > 0
2 * 2 * 2 * ... * 2 = 2^x > 0
You can prove by inducton that 2^x > 0 for all x in R:
We know that 2 > 0.
Assume 2^k > 0.
Then 2^(k+1) = 2 * (2^k). When you multiply two positive numbers, you get a positive number. 2 is positive. 2^k is positive. Hence 2 * (2^k) is positive. End of proof.
This obviously works for any positive number A, but you can prove it the same way by writing A instead of 2 and knowing that A is positive.
Now, instead of 2, let A = e = 2.71828...
Obviously e^x will also be > 0 for all x in R.
If you were working with complex numbers, then e^x can be negative. For example, e^i? = -1. But clearly you're working with reals, so you don't need that.
>2x draws a negative line for x<0, and a positive line for x>0.Just solve the equation bro:
>when is 2x negative?2x < 0
x < 0
>so, when x < 0, then 2x is negative.Do the same for x>0.
How old are you OP?
