Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.
f(x) = 4 – x²; f′(−3), f′(0), f′(1)
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Start by recalling the definition of the derivative: f'(x) = lim(h -> 0) [(f(x + h) - f(x)) / h]. This will be used to find the derivative of the function f(x) = 4 - x².
Factor out h from the numerator: f'(x) = lim(h -> 0) [-h(2x + h) / h]. Cancel h from numerator and denominator, then evaluate the limit as h approaches 0 to find f'(x) = -2x. Finally, substitute x = -3, 0, and 1 into f'(x) to find f'(-3), f'(0), and f'(1).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definition of Derivative
The derivative of a function at a point is the limit of the average rate of change of the function as the interval approaches zero. Mathematically, it is defined as f'(x) = lim(h→0) [(f(x+h) - f(x))/h]. This concept is essential for understanding how to calculate the instantaneous rate of change of a function.
The power rule is a basic derivative rule used to find the derivative of functions in the form of x^n. It states that if f(x) = x^n, then f'(x) = n*x^(n-1). This rule simplifies the process of finding derivatives for polynomial functions, such as f(x) = 4 - x², where the derivative is calculated using this rule.
Once the derivative function is found, it can be evaluated at specific points to find the slope of the tangent line at those points. For example, after finding f'(x) for f(x) = 4 - x², you can substitute x = -3, 0, and 1 into f'(x) to find f'(-3), f'(0), and f'(1), respectively, which represent the instantaneous rate of change at these points.