cx+7 if x<or=2
f(x)= cx^2-5 if x>2
determine any values of c which make f continuous for all real numbers
(c=6)
determine any values of c which make f differentiable for all real numbers
???
To determine the values of c that make f continuous for all real numbers, we need to ensure that the function is continuous at x = 2.
For continuity at x = 2, the limit of the function as x approaches 2 from the left (x <= 2) should be equal to the limit of the function as x approaches 2 from the right (x > 2).
Let's evaluate the limits:
Limit as x approaches 2 from the left (x <= 2):
lim(x->2-) f(x) = lim(x->2-) (cx + 7)
= c(2) + 7
= 2c + 7
Limit as x approaches 2 from the right (x > 2):
lim(x->2+) f(x) = lim(x->2+) (cx^2 - 5)
= c(2^2) - 5
= 4c - 5
For continuity, these two limits should be equal. Therefore:
2c + 7 = 4c - 5
Solving this equation, we get:
2c - 4c = -5 - 7
-2c = -12
c = 6
So, the value of c that makes f continuous for all real numbers is c = 6.
Now, let's determine the values of c that make f differentiable for all real numbers. For differentiability, the function needs to be continuous, and the derivative of the function should exist at every point.
We already determined that c = 6 makes f continuous. To check differentiability, let's differentiate the function f(x) for x > 2:
f'(x) = d/dx (cx^2 - 5)
= 2cx
The derivative exists for all real numbers when c is any real number, including the case when c = 6.
Therefore, any value of c (including c = 6) makes the function f(x) differentiable for all real numbers.