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?
f(x) is both functions.. I don't know why the system didn't let me keep the spaces to show that
First off, stop misspelling "calculus".
Now, if a=2 and b=3, we have
f(x) = 2x^2 - 3x
f(1) = -1
But lim x->1- = |1-1| + 2 = 2
So, f is not continuous at x=1.
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If f is to be continuous, it needs to be continuous at x=1. It is continuous everywhere else already.
So, ax^2 - bx must = 2 at x=1
a - b = 2
So, there are any number of parabolas which will make f continuous at x=1.
6x^2 - 4x
-3x^2 + 5x
etc.
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Now, for f to be also differentiable, the slopes must match at x=1
The slope of |x-1| is -1 when approaching from the left.
ax^2 - (a-2)x must also have slope -1 at x=1
f'(x) = 2ax - (a-2)
f'(1) = 2a - a + 2 = -1
a = -3
So,
f(x) = |x-1| + 2 for x<1
f(x) = -3x^2 + 5x for x>=1
is both continuous and differentiable everywhere.
a) To determine if f is continuous for all x, we need to check if the left-hand limit at x = 1 is equal to the right-hand limit at x = 1, and also if the function value at x = 1 is equal to these limits.
Left-hand limit as x approaches 1: lim(x→1-) |x-1| + 2 = |1-1| + 2 = 2
Right-hand limit as x approaches 1: lim(x→1+) ax^2 - bx = (2)(1)^2 - (3)(1) = 2 - 3 = -1
Function value at x = 1: f(1) = (2)(1)^2 - (3)(1) = 2 - 3 = -1
Since the left-hand limit, right-hand limit, and function value at x = 1 are not equal (-1 ≠ 2), f is not continuous for all x when a = 2 and b = 3.
b) To find the values of a and b for which f is a continuous function, we need to ensure that the left-hand limit, right-hand limit, and function value at x = 1 are equal.
Left-hand limit as x approaches 1: lim(x→1-) |x-1| + 2 = |1-1| + 2 = 2
Right-hand limit as x approaches 1: lim(x→1+) ax^2 - bx = a(1)^2 - b(1) = a - b
Function value at x = 1: f(1) = a(1)^2 - b(1) = a - b
Therefore, for f to be continuous, we must have 2 = a - b.
c) To find the values of a and b for which f is both continuous and differentiable, we need to ensure that the left-hand derivative and the right-hand derivative at x = 1 exist and are equal.
Left-hand derivative at x = 1: lim(x→1-) [f(x) - f(1)] / (x - 1) = lim(x→1-) [(|x-1| + 2) - (a - b)] / (x - 1) = 1
Right-hand derivative at x = 1: lim(x→1+) [f(x) - f(1)] / (x - 1) = lim(x→1+) [(ax^2 - bx) - (a - b)] / (x - 1) = 2a - b
Therefore, for f to be both continuous and differentiable, we must have 1 = 2a - b.
Combining both conditions, we have the system of equations:
2 = a - b
1 = 2a - b
Solving this system of equations will give us the values of a and b for which f is both continuous and differentiable.
a) To determine if f is continuous for all x, we need to check if the left and right limits of f(x) at x = 1 are equal to f(1).
For x < 1:
When x approaches 1 from the left (x → 1-), the function becomes |1-1| + 2 = 2.
For x ≥ 1:
When x approaches 1 from the right (x → 1+), the function becomes a(1)^2 - b(1) = a - b.
To determine if f is continuous at x = 1, we need the left limit, right limit, and the function value at x = 1 to be equal.
From the given conditions, we have:
Left limit (x → 1-) = 2
Right limit (x → 1+) = a - b
Function value (f(1)) = a(1)^2 - b(1) = a - b
For f to be continuous at x = 1, the left and right limits of f(x) and the function value at x = 1 need to be equal. So, we equate them:
2 = a - b (Equation 1)
If a = 2 and b = 3, we can substitute these values into Equation 1:
2 = (2) - (3)
2 = 2 - 3
2 = -1
Since the equation does not hold true, f is not continuous for all x when a = 2 and b = 3.
b) Now we need to find the values of a and b for which f is a continuous function. Referencing Equation 1 from part a), we have:
2 = a - b
To make f continuous, we need a and b to satisfy this equation. However, we can also consider that for f to be continuous, both the left and right limits as x approaches 1 must exist and be equal. This means that the value of a does not matter as long as b = a - 2.
So, the values of a and b for which f is a continuous function are:
a can be any value.
b = a - 2.
c) To determine the values of a and b for which f is both continuous and differentiable, we need to ensure that the derivative of f exists and is continuous at x = 1.
For x < 1:
The derivative of |x - 1| + 2 is 1 if x < 1.
For x ≥ 1:
The derivative of ax^2 - bx is 2ax - b.
To ensure f is differentiable at x = 1, the left and right derivatives must exist and be equal.
From the left (x → 1-): The derivative is 1.
From the right (x → 1+): The derivative is 2a - b.
To find the values of a and b for which f is continuous and differentiable, we need to equate the left and right derivatives.
1 = 2a - b
Using this equation, we can find the values of a and b which satisfy both continuity and differentiability conditions.