Suppose f(x)=x2−1 for x≤1 and f(x)=acosbx for x>1 . What values of a and b will ensure that f(x) is differentiable for all values of x? (1 point) Responses a=−4π, b=π4 a = − 4 π , b = π 4 a=−4π, b=π2 a = − 4 π , b = π 2 a=−2π, b=π4 a = − 2 π , b = π 4 a=−2π, b=π2 a = − 2 π , b = π 2 a=2π, b=π2
wrong again.
having the derivatives match is not all that is required. We must also have f(x) be continuous at x=1.
Since x^2-1 = 0, we need a cos(bx) = a cos(b) = 0
That is, b must be an odd multiple of π/2
Since we have sin(b) = -2/ab, sin(b) will be either 1 or -1
if b = π/2, 1 = -2/(a*π/2) so a = -4/π
if b = -π/2, -1 = -2/(a(-π/2)) so a = -4/π
To ensure that f(x) is differentiable for all values of x, we need the two parts of the function to smoothly connect at x = 1, meaning their derivatives must match at that point.
Taking the derivative of f(x) = x^2 - 1 for x ≤ 1:
f'(x) = 2x
Taking the derivative of f(x) = acosbx for x > 1:
f'(x) = -ab*sin(bx)
Setting these two derivatives equal to each other at x = 1 and solving for a and b:
2(1) = -ab*sin(b)
2 = -ab*sin(b)
To satisfy this equation, we need a and b to be such that sin(b) = -2/ab. The only option among the given answer choices is a=-4π and b=π/4, as the others do not yield a solution for sin(b) that satisfies the equation. Therefore, the values a = -4π and b = π/4 will ensure that f(x) is differentiable for all values of x.
You are correct. I apologize for the mistake in my previous response.
To ensure that f(x) is differentiable for all values of x, we need the two parts of the function to smoothly connect at x = 1, meaning their values must be equal at that point.
For x ≤ 1, we have f(x) = x^2 - 1. At x = 1, f(x) = 1^2 - 1 = 0.
For x > 1, we have f(x) = acos(bx). At x = 1, f(x) = acos(b).
To ensure continuity at x = 1, we need acos(b) = 0. This implies that either a = 0 or cos(b) = 0.
If a = 0, then f(x) = 0 for all x > 1, which means it is a constant function and automatically differentiable.
If cos(b) = 0, then b must be an odd multiple of π/2.
Therefore, the values a = 0 and b = π/2 or a = 0 and b = -π/2 will ensure that f(x) is differentiable for all values of x.