The solution for this problem would be:
f'(x) = 1 - 1/x^2 = (x^2 - 1)/x^2
is positive where |x| > 1
hence (-inf, -1) and (1, inf) are the regions in which f' is positive and therefore f is increasing.
Therefore, the answer is (-infinity,-1] U [1,infinity).
Answer:
See solutions below
Step-by-step explanation:
For what values of X are the statements below true
A. 1x>x+1
x >x+1
x-x>1
0x > 1
x > 1/0
X >∞
B) |1-x|>3
The fucntion can both be positive and negative
For the negative function
-(1-x) > 3
-1+x > 3
x > 3+1
x > 4
For the positive function
1-x > 3
-x > 3 - 1
-x > 2
x < -2
Hence the required solutions are x > 4 and x < -2
c) For the equation
|x-15| < 0
-(x - 15) < 0
-x + 15 < 0
-x < -15
x > 15
Also x-15 < 0
x < 0+15
x < 15
Hence the required solution is x > 15 and x < 15
Answer: 5
Step-by-step explanation:
Answer:
12
Step-by-step explanation:
This is assuming that x2 means x to the power of 2 and that the x on the left is to represent multiplying. First, plug in all of the number to get 2*3+3^2-3.Using PEMDAS, you should solve the exponent to get 9. Then, multiply 2 by 3 to get 6. Add 6 to 9 to get 15. Subtract 2 to 15 to get 12.
TL;DR
2*3+3^2-3
6+9-3
15-3
12
