Let f(x) = x². a. Show that f(x)−f(y) / x−y = f′(x+y²), for all x≠y.
Verified step by step guidance
1
Start by understanding the problem: We need to show that the difference quotient (f(x) - f(y)) / (x - y) is equal to the derivative of f evaluated at (x + y²).
First, calculate f(x) and f(y) using the given function f(x) = x². This gives us f(x) = x² and f(y) = y².
Substitute these into the difference quotient: (f(x) - f(y)) / (x - y) = (x² - y²) / (x - y).
Recognize that x² - y² is a difference of squares, which can be factored as (x - y)(x + y). Substitute this into the difference quotient to simplify: ((x - y)(x + y)) / (x - y).
Cancel the (x - y) terms in the numerator and denominator, assuming x ≠ y, to get x + y. Now, find the derivative f'(x) = 2x, and evaluate it at x + y²: f'(x + y²) = 2(x + y²). Compare this with the simplified expression x + y to verify the equality.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2m
Play a video:
0 Comments
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Difference Quotient
The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is defined as (f(x) - f(y)) / (x - y) for x ≠ y. This expression is crucial for understanding the derivative, as it approaches the instantaneous rate of change as y approaches x.
The derivative of a function at a point measures how the function's output changes as its input changes. It is denoted as f'(x) and can be interpreted as the slope of the tangent line to the function's graph at that point. In this question, we need to show that the difference quotient equals the derivative of f at a specific point, which involves applying the definition of the derivative.
Function composition involves combining two functions to create a new function, where the output of one function becomes the input of another. In this context, we need to evaluate f' at (x + y²), which requires understanding how to apply the derivative to a function that is itself a composition of variables. This concept is essential for manipulating and simplifying expressions involving derivatives.