Derivatives and tangent lines a. For the following functions and values of a, find f′(a). f(x) = 1/3x-1; a= 2
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Step 1: Identify the function f(x) = \(\frac{1}{3}\)x - 1 and the point a = 2 where you need to find the derivative f'(a).
Step 2: Recall that the derivative f'(x) of a linear function f(x) = mx + b is simply the coefficient m of x. In this case, f(x) = \(\frac{1}{3}\)x - 1, so the derivative f'(x) is \(\frac{1}{3}\).
Step 3: Since the derivative of a linear function is constant, f'(x) = \(\frac{1}{3}\) for all x. Therefore, f'(a) = f'(2) = \(\frac{1}{3}\).
Step 4: Interpret the result: The derivative f'(a) = \(\frac{1}{3}\) represents the slope of the tangent line to the graph of f(x) at the point where x = 2.
Step 5: Conclude that the slope of the tangent line at x = 2 is \(\frac{1}{3}\), which means the line rises \(\frac{1}{3}\) unit for every 1 unit it moves to the right.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivatives
A derivative represents the rate of change of a function with respect to its variable. It is defined as the limit of the average rate of change of the function as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the function at that point.
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. This concept is crucial for understanding how functions behave locally around specific values.
Function evaluation involves substituting a specific value into a function to determine its output. In the context of derivatives, evaluating the function at a point 'a' allows us to find the slope of the tangent line at that point, which is essential for understanding the function's behavior at that specific location.