Let f be he real-valued function defined by f(x)=sin^3x+sin^3|x|
(a) find f'(x) for x>0
(b) find f'(x) for x<0
(c) determine whether f(x) is continuous and x=0
(d) determine whether the derivative of f(x) exists at x=0
(a) Well, well, let's take a look at f'(x) for x > 0. To find the derivative of f(x), we need to use the chain rule since we have this absolute value business going on. Let's start by differentiating sin^3x, which is 3sin^2x * cosx. The derivative of sin^3|x| will be the same, because absolutes are just sneaky little creatures that don't change much. So, f'(x) for x > 0 is 3sin^2x * cosx + 3sin^2|x| * cos|x|.
(b) Now, for x < 0, we again have to apply the chain rule to tackle the absolute value. The derivative of sin^3x will be the same as before, 3sin^2x * cosx. However, the derivative of sin^3|x| will be a bit different. We have to remember that the inside of the absolute value is negative, so the derivative will be -3sin^2|x| * cos|x|. Therefore, f'(x) for x < 0 is 3sin^2x * cosx - 3sin^2|x| * cos|x|.
(c) Is f(x) continuous at x = 0? Well, to determine that, we need to check if f(0) exists and if the limit of f(x) as x approaches 0 exists, and if they are the same. Let's plug in x = 0 into f(x) and see what happens. We get f(0) = sin^3(0) + sin^3|0|, which simplifies to 0 + 0 = 0. So, f(0) does exist.
Now, let's take the limit as x approaches 0. As x approaches 0 from the positive side, sin^3x approaches 0, and as x approaches 0 from the negative side, sin^3|x| also approaches 0. Therefore, the limit of f(x) as x approaches 0 is also 0. Since the value of f(x) and the limit agree, we can conclude that f(x) is continuous at x = 0.
(d) Now, this one's a bit trickier. Does the derivative of f(x) exist at x = 0? To determine that, we need to check if the left-handed derivative and the right-handed derivative are the same. So, we need to calculate f'(x) for x > 0 and x < 0, and see if they match at x = 0.
From part (a), we had f'(x) for x > 0 as 3sin^2x * cosx + 3sin^2|x| * cos|x|. From part (b), we had f'(x) for x < 0 as 3sin^2x * cosx - 3sin^2|x| * cos|x|. Now, let's evaluate them at x = 0.
For x > 0: f'(0) = 3sin^2(0) * cos(0) + 3sin^2|0| * cos|0| = 0 + 0 = 0.
For x < 0: f'(0) = 3sin^2(0) * cos(0) - 3sin^2|0| * cos|0| = 0 - 0 = 0.
Since both the left-handed and right-handed derivatives are equal at x = 0, we can conclude that the derivative of f(x) does exist at x = 0. Phew, mission accomplished!
(a) To find f'(x) for x > 0, we can differentiate each term separately using the chain rule. Let's start with the first term sin^3x:
Using the chain rule, we get:
d/dx [sin^3x] = 3sin^2x * cosx
Now let's differentiate the second term sin^3|x|:
Using the chain rule, we need to consider two cases: when x > 0 and x < 0.
For x > 0:
d/dx [sin^3|x|] = d/dx [sin^3x] = 3sin^2x * cosx
For x < 0:
d/dx [sin^3|x|] = d/dx [-sin^3x] = -3sin^2x * cosx
To find f'(x) for x > 0, we add the derivatives of the two terms together:
f'(x) = 3sin^2x * cosx + 3sin^2x * cosx = 6sin^2x * cosx
(b) To find f'(x) for x < 0, we add the derivatives of the two terms with opposite signs:
f'(x) = 3sin^2x * cosx - 3sin^2x * cosx = 0
(c) To determine whether f(x) is continuous at x = 0, we need to evaluate the limit of f(x) as x approaches 0 from both the left and the right.
For x > 0:
lim(x → 0+) [sin^3x + sin^3|x|] = sin^30 + sin^30 = 0 + 0 = 0
For x < 0:
lim(x → 0-) [sin^3x + sin^3|x|] = sin^30 + sin^30 = 0 + 0 = 0
Since the limit of f(x) as x approaches 0 from both sides is 0, we can conclude that f(x) is continuous at x = 0.
(d) To determine whether the derivative of f(x) exists at x = 0, we need to check if the left-hand derivative and the right-hand derivative at x = 0 are equal.
For x > 0:
f'(0+) = lim(x → 0+) [f'(x)] = lim(x → 0+) [6sin^2x * cosx] = 6sin^20 * cos0 = 0
For x < 0:
f'(0-) = lim(x → 0-) [f'(x)] = lim(x → 0-) [0] = 0
Since the left-hand derivative and right-hand derivative are equal at x = 0, we can conclude that the derivative of f(x) exists at x = 0, and it is equal to 0.
To find the derivative of the function f(x) = sin^3(x) + sin^3(|x|) for different values of x, let's break it down into three cases: x > 0, x < 0, and x = 0.
(a) Derivative of f(x) for x > 0:
For x > 0, the function f(x) = sin^3(x) + sin^3(x) since |x| = x. To find the derivative, we'll differentiate each term individually using the chain rule.
First, let's differentiate sin^3(x):
Using the chain rule, the derivative of sin^3(x) is 3sin^2(x)cos(x).
Next, let's differentiate sin^3(x) again:
The derivative of sin^2(x)cos(x) is 2sin(x)cos^2(x) - sin^3(x).
Therefore, the derivative of f(x) = sin^3(x) + sin^3(x) is f'(x) = 3sin^2(x)cos(x) + 2sin(x)cos^2(x) - sin^3(x).
(b) Derivative of f(x) for x < 0:
For x < 0, the function f(x) = sin^3(x) + sin^3(-x) since |x| = -x. Now, let's differentiate this function.
Differentiating sin^3(x) with respect to x gives us the same result as in part (a), which is 3sin^2(x)cos(x).
Now, let's differentiate sin^3(-x):
Using the chain rule, the derivative of sin^3(-x) is -3sin^2(-x)cos(-x).
Remember that sin(-x) = -sin(x) and cos(-x) = cos(x).
Therefore, the derivative of f(x) = sin^3(x) + sin^3(-x) is f'(x) = 3sin^2(x)cos(x) - 3sin^2(x)cos(x) = 0.
(c) Continuity of f(x) at x = 0:
To determine if f(x) is continuous at x = 0, we need to check if the left-hand limit, right-hand limit, and the function value at x = 0 are equal.
The left-hand limit as x approaches 0 is f(0-) = sin^3(0) + sin^3(0) = 0 + 0 = 0.
The right-hand limit as x approaches 0 is f(0+) = sin^3(0) + sin^3(0) = 0 + 0 = 0.
The function value at x = 0 is f(0) = sin^3(0) + sin^3(0) = 0 + 0 = 0.
Since the left-hand limit, right-hand limit, and the function value at x = 0 are all equal to 0, f(x) is continuous at x = 0.
(d) Existence of derivative at x = 0:
To determine if the derivative of f(x) exists at x = 0, we need to check if the left-hand derivative and right-hand derivative are equal.
The left-hand derivative is f'(0-) = 0 since we obtained a derivative of 0 for x < 0 in part (b).
The right-hand derivative is f'(0+) = 0 since we obtained a derivative of 0 for x > 0 in part (a).
Since the left-hand derivative and the right-hand derivative are both 0, the derivative of f(x) exists at x = 0.