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stl11
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f(x)=xabs(x) show f '(0) exists
The function f(x) = x| x | represents a piecewise function that takes the absolute value of x and multiplies it by x. This means that the function will have a "V" shape with the point at (0,0) and the left and right branches having a slope of 1 and -1 respectively.
To determine if the derivative of f(x) exists at x = 0, we need to take the limit of the function as x approaches 0 from both the left and right sides. If the left and right limits are equal, then the derivative exists at x = 0. In this case, the derivative of f(x) at x = 0 is 0.
The formula for finding the derivative of f(x) = x| x | is f'(x) = 2x if x > 0 and f'(x) = -2x if x < 0. This means that the derivative is a piecewise function with a slope of 2 on the left side of the "V" and a slope of -2 on the right side of the "V". At x = 0, the derivative is 0.
Yes, the function f(x) = x| x | is differentiable at x = 0. As mentioned in the previous questions, the left and right limits of the function at x = 0 are equal, meaning that the derivative exists at this point and is equal to 0.
Yes, the function f(x) = x| x | can be extended to become a continuous function by assigning a value of 0 at x = 0. This means that the function will have a "V" shape with a point at (0,0) and a connected line on either side with a slope of 1 and -1 respectively. This extended function will be continuous at all points, including x = 0.