Answer:
Observe that f(x) is a continuous function when
because is a polynomial. The possible problem may occur in x=1.
Then, f(x) is discontinuous in x=1 if the limits of f to the right and the left of 1 exist and are different or if some of those limits doesn't exist.
Let's calculate the limits:


Since,
then f(x) is discontinuous in x=1.
For a quadratic equation ax^2 + bx + c = 0, the discriminant is given by b^2 - 4ac
Thus for a^2 - 2a + 5 = 0,
a = 1, b = -2 and c = 5
b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16
Answer:
283.495
Step-by-step explanation: