let f be the function defined by
|x-1|+2 for X<1
f(x)=
ax^2-Bx, for X>or equal to 1. where a and b are constants
a)if a=2 and b=3 is f continious for all x? justify your answer
b)describe all the values of a and b for which f is a continious function
c) For what values of a and b is f both continious and differentiable?
a) To determine if a function is continuous, we need to check if it is continuous at every point in its domain. In this case, we have two different expressions for f(x) depending on the value of x.
When x < 1, the function is given by |x-1|+2.
When x >= 1, the function is given by ax^2 - bx.
For f to be continuous for all x, we need the two expressions to match up when x = 1. That means, we need to find the values of a and b such that:
|x-1| + 2 = ax^2 - bx ... (Equation 1)
Let's substitute x = 1 into Equation 1:
|1-1| + 2 = a(1^2) - b(1)
2 = a - b ... (Equation 2)
So, for f to be continuous for all x, we need a - b = 2.
b) To determine the values of a and b for which f is a continuous function, we can look at the conditions derived in part a. We found that for f to be continuous, we need a - b = 2. This means any values of a and b that satisfy this condition will make f a continuous function.
c) For f to be both continuous and differentiable, it must be continuous at every point in its domain, and the derivatives of the two expressions must match up when x = 1.
For the first expression |x-1| + 2, the derivative is 1 when x < 1.
For the second expression ax^2 - bx, the derivative is 2ax - b when x >= 1.
To make f both continuous and differentiable, we need the derivatives of the two expressions to match up when x = 1. So we have:
1 = 2a - b ... (Equation 3)
In addition, the condition derived in part a, a - b = 2, must also be satisfied.
Solving Equations 2 and 3 simultaneously, we have:
a - b = 2
2a - b = 1
Solving these equations, we get a = 3/2 and b = 1/2.
Therefore, for f to be both continuous and differentiable, the values of a and b must be a = 3/2 and b = 1/2.