Condition 2: f(x) exists.Condition 3: f(x) = f(c)Notice that the function is not defined at. Therefore, f(x) is not continuous at x = 3 because it
fails condition 1.- The function is continuous for k = 6 or k = −1.
In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the
following conditions:Condition 1: f(c) exists.Condition 2: f(c) exists.Condition 3: f(c) = f(c)We will need to find a value, or values, of k that enables f(x) to satisfy each condition.Condition 1: f(−3) = k^2 − 5kCondition 2: f(x) = 6 and f(x) = 6, so f(x) = 6.Condition 3: Now we need to find a value, or values, of k such that f(x) = 6 = f(3). If weset k^2 − 5k = 6, we obtain the solutions k = 6 and k = −1.11. 20
We find the derivative of a function, f(x), using the definition of the derivative, which is: f′(x) =. Here f(x) = 2x^2 and x = 5. This means that f(5) = 2(5)^2 = 50 and f(5 +
h) = 2(5 + h)^2 = 2(25 + 10h + h^2 ) = 50 + 20h + 2h^2 . If we now plug these into the definition ofthe derivative, we get f′(5) = = . Thissimplifies to f′(5) = . Now we can factor out the h from the numerator andcancel it with the h in the denominator: f′(5) = = (20 + 2h). Now wetake the limit to get f′(5) = (20 + 2h) = 20.