By the confront theorem we know that the limit only exists if both lateral limits are equal
In this case they aren't so we don't have limit for x approaching 2, but we can find their laterals.
Approaching 2 by the left we have it on the 5 line so this limit is 5
Approaching 2 by the right we have it on the -3 line so this limit is -3
Think: it's approaching x = 2 BUT IT'S NOT 2, and we only have a different value for x = 2 which is 1, but when it's approach by the left we have the values in the 5 line and by the right in the -3 line.
f(x) = log2 x
f(40) = log2 40
40 = 2^y
2^5 = 32 and 2^6 = 64
so f(40) lies between integers 5 and 6.
Answer:
4x^2 + 4x -2
Step-by-step explanation:
Use FOIL to solve (x+2) (x+2)
4(x^2 + 4x +4) - 6
4x^2 + 4x -2
f(-9) means x is -9
Replace x with -9 and solve:
2/3(-9) - 6
Simplify:
-6 - 6 = -12
f(-9) = -12
