There are no values of x for which f'(x)=−3.
Let's tackle each part of the problem:
(a) Show that f is continuous at x 0
To show that f is continuous at x = 0, we need to check that f(x) approaches the same value from both sides as approaches 0.
For x ≤ 0, f(x) = 1 - 2 sin(x).
For x > 0, f(x) = -4x.
Now, let's find the limit as 2 approaches 0 from both sides:
limo f(x) = limx→0 (1-2 sin(x)) = 1
limx→0+ f(x) = limx→0+(-4x) = 0
Since limx→0 f(x) = limx→0+ f(x), the limit exists and is the same from both sides, and thus, f is continuous at x = 0.
(b) Express f'(x) as a piecewise-defined function and find x for which f'(x) = -3.
Let's find the derivative f'(x) for x≠ 0:
For x < 0, f(x) = 1 - 2 sin(x). The derivative is:
f'(x)=-2 cos(x)
For x > 0, f(x) = -42. The derivative is:
f'(x) = -4
So, the piecewise-defined derivative is:
Alan & Stanley D
Now, set f'(x) = -3 and solve for x:
For x ≤ 0:-2 cos(x) = -3
cos(x) =2/3
The cosine function is bounded between -1 and 1, so there are no solutions for x≤0.
For x>0: −4=−3 This equation has no solution.
So, there are no values of x for which f'(x)=−3.