f'(x) if x is less than or greater than 1
Let F'(x)={
g'(x) if x>1
Where f'(x)=x/(x^2-2)^2 and g'(x)=3x-4
A. If f(1)=3/2, find f(x)
B. If F(x) is continuous at x=1, find g(x).
C. Express F(x).
D. If F(x) is continuous at x=1, determine if F(x) is differentiable at x=1.
It's supposed to be a piecewise function. And I have no idea how to go about this problem. Can you please help? Thanks.
Certainly, I'd be happy to help you solve this problem step by step. Let's break it down:
A. To find f(x) given f'(x) and f(1), we need to integrate f'(x) with respect to x. However, since f'(x) is defined for x less than or greater than 1, we need to separate the integration into two intervals: x < 1 and x > 1.
For x < 1:
Since f'(x) = x/(x^2-2)^2, we can integrate f'(x) with respect to x to find f(x). In this case, we can use u-substitution to simplify the integration. Let u = x^2 - 2, then du = 2x dx.
So, the integral for x < 1 becomes:
∫[x < 1] f'(x) dx = ∫[x < 1] (x/(x^2-2)^2) dx = (1/2) ∫(u^-2) du
Now, integrating (u^-2) with respect to u gives us:
∫(u^-2) du = -u^-1 + C = -1/(x^2 - 2) + C
For x > 1:
Since f'(x) = g'(x), we can use g'(x) to find f(x). So, in this interval, f(x) = g(x).
Therefore, for x > 1:
f(x) = g(x) = ∫[x > 1] g'(x) dx = ∫[x > 1] (3x - 4) dx = (3/2)x^2 - 4x + C
Now that we have expressions for f(x) in both intervals, we can write the piecewise function for f(x):
f(x) = {
-1/(x^2 - 2) + C for x < 1
(3/2)x^2 - 4x + C for x > 1
B. To find g(x) given F(x) is continuous at x = 1, we need to evaluate the limit of f(x) as x approaches 1 from both sides:
lim[x->1^-] f(x) = lim[x->1^-] (-1/(x^2 - 2) + C)
Since f(x) is continuous at x = 1, we can evaluate this limit.
lim[x->1^+] f(x) = lim[x->1^+] ((3/2)x^2 - 4x + C)
Since f(x) is continuous at x = 1, we can evaluate this limit as well.
Now, since lim[x->1^-] f(x) = lim[x->1^+] f(x), we can equate the expressions:
-1/(1^2 - 2) + C = (3/2)(1^2) - 4(1) + C
Simplifying the equation:
-1/(-1) + C = 3/2 - 4 + C
1 + C = -4/2 + C
1 + C = -2 + C
By cancelling C from both sides, 1 = -2 which is not possible.
Therefore, the function is not continuous at x = 1.
C. To express F(x), we need to combine the expressions for f(x) in both intervals:
F(x) = {
-1/(x^2 - 2) + C for x < 1
(3/2)x^2 - 4x + C for x > 1
D. Since we determined that F(x) is not continuous at x = 1, it is not differentiable at x = 1. The definition of differentiability requires both continuity and the existence of the derivative at a point. Since F(x) fails to be continuous at x = 1, it cannot be differentiable at that point.
I hope this explanation helps! Let me know if you have any further questions.