Q. For the function f(x)=∣3x−9∣, evaluate the left and right limits of f′(x) as x approaches 2 .a. limx→2−f(x)3b. limx→2+f(x)3
Understand Absolute Value Function: To find the left and right limits of f′(x) as x approaches 2, we first need to understand the behavior of the absolute value function f(x)=∣3x−9∣. The absolute value function has a kink at the point where the expression inside the absolute value is zero. In this case, that point is when 3x−9=0, which occurs at x=3. Since we are interested in the limits as x approaches 2, we are looking at the behavior of the function to the left and right of x=2, but still to the left of the kink at x=3.
Analyze Function Behavior: For x<3, the expression inside the absolute value, 3x−9, is negative. Therefore, the function f(x) for x<3 is f(x)=−(3x−9)=−3x+9. The derivative of this function with respect to x is f′(x)=−3.
Calculate Derivative: For x>3, the expression inside the absolute value, 3x−9, is positive. Therefore, the function f(x) for x>3 is f(x)=3x−9. The derivative of this function with respect to x is f′(x)=3.
Determine Left and Right Limits: Since we are interested in the limits as x approaches 2, we only need to consider the behavior of the function for x<3. Therefore, both the left-hand limit and the right-hand limit of f′(x) as x approaches 2 are equal to the derivative of the function for x<3, which is −3.
Calculate Left-Hand Limit: The left-hand limit of f′(x) as x approaches 2 is: limx→2−f′(x)=−3
Calculate Right-Hand Limit: The right-hand limit of f′(x) as x approaches 2 is: limx→2+f′(x)=−3