Answer:
For f(x) to be differentiable at 2, k = 5.
Step-by-step explanation:
For f(x) to be differentiable at x = 2, f(x) has to be continuous at 2.
For f(x) to be continuous at 2, the limit of f(2 – h) = f(2) = f(2 + h) as h tends to 0.
Now,
f(2 – h) = 2(2 – h) + 1 = 4 – 2h + 1 = 5 – 2h.
As h tends to 0, lim (5 – 2h) = 5
Also
f(2 + h) = 3(2 + h) – 1 = 6 + 3h – 1 = 5 + 3h
As h tends to 0, lim (5 + 3h) = 5.
So, for f(2) to be continuous k = 5
Tom is incorrect
He keeps getting the same number of 1:6 on each roll.
Hello!
Your answer will be $150! Hope this helps! :D
Answer:
Step-by-step explanation:
The answer to the question is: Yes.
Explanation: It can be solved by following exactly the same procedures
that solved the other two questions that you posted within the past 15 minutes.
Since I'm here already, here is the solution to this one:
6x² + 18x = 0
Divide each side of the equation by 6 :
x² + 3x = 0
Factor the left side:
x(x+ 3) = 0
This statement is true if x=0 or if x=3.
The solution to this one is even the same as the solution to the last one you posted.
You really ought to take a break, go back, review the solutions you've been given,
and try to solve a few on your own.
