Function Evaluation And Difference Quotient For F(x) = 2x² + 5
In mathematics, understanding functions is crucial. Functions are mathematical relationships that map inputs to outputs. Evaluating a function at a specific point involves substituting that point into the function's expression and simplifying. The difference quotient, on the other hand, is a fundamental concept in calculus that helps us understand the rate of change of a function. In this comprehensive guide, we will explore how to find f(a) and f(a+h) for the function f(x) = 2x² + 5, and then delve into calculating the difference quotient (f(a+h) - f(a))/h. This exploration will not only solidify your understanding of function evaluation but also introduce you to the essential concept of the difference quotient, a cornerstone of differential calculus.
To evaluate f(a), we substitute 'a' for 'x' in the function f(x) = 2x² + 5. This process replaces every instance of the variable x with the value a. This is a direct application of the function's definition, where we are finding the output of the function when the input is a. Understanding this substitution process is fundamental to working with functions and forms the basis for more complex operations. The result of this substitution gives us a new expression in terms of a, which represents the function's value at the point a. Performing this substitution correctly is vital for further calculations and analysis. Let's proceed with the substitution:
f(a) = 2(a)² + 5
f(a) = 2a² + 5
This result, 2a² + 5, is the value of the function f(x) when x is equal to a. It's a quadratic expression in a, reflecting the quadratic nature of the original function. Understanding how to perform this substitution is a foundational skill in algebra and calculus, and it allows us to analyze the behavior of functions at specific points.
Next, we evaluate f(a + h). This step involves substituting (a + h) for x in the function f(x) = 2x² + 5. This is a crucial step in understanding how the function behaves when its input is slightly perturbed by a value h. The substitution requires careful attention to algebraic manipulation, particularly when dealing with squared terms. We'll need to expand the squared term (a + h)² correctly to ensure we obtain the correct expression for f(a + h). This process will reveal how the function's output changes as the input varies from a to a + h, which is a key concept in understanding rates of change and derivatives. So, let's substitute and simplify:
f(a + h) = 2(a + h)² + 5
Now, we expand the squared term:
f(a + h) = 2(a² + 2ah + h²) + 5
Distribute the 2:
f(a + h) = 2a² + 4ah + 2h² + 5
This expression, 2a² + 4ah + 2h² + 5, represents the value of the function f(x) when x is equal to (a + h). It contains terms involving a², ah, and h², reflecting the quadratic nature of the function and the shift in input by h. This result is essential for calculating the difference quotient, which we'll explore next.
The difference quotient, (f(a + h) - f(a))/h, is a central concept in calculus. It represents the average rate of change of the function f(x) over the interval from a to a + h. In essence, it's the slope of the secant line connecting the points (a, f(a)) and (a + h, f(a + h)) on the graph of the function. This concept is a precursor to the derivative, which is the instantaneous rate of change and is found by taking the limit of the difference quotient as h approaches 0. Calculating the difference quotient involves subtracting f(a) from f(a + h), dividing the result by h, and then simplifying the expression. This process often involves algebraic manipulation to cancel out terms and simplify the quotient. Let's calculate the difference quotient for our function:
We have:
f(a + h) = 2a² + 4ah + 2h² + 5
f(a) = 2a² + 5
Now, we compute f(a + h) - f(a):
f(a + h) - f(a) = (2a² + 4ah + 2h² + 5) - (2a² + 5)
f(a + h) - f(a) = 2a² + 4ah + 2h² + 5 - 2a² - 5
f(a + h) - f(a) = 4ah + 2h²
Next, we divide by h:
(f(a + h) - f(a))/h = (4ah + 2h²)/h
We can factor out an h from the numerator:
(f(a + h) - f(a))/h = h(4a + 2h)/h
Now, we cancel the h in the numerator and denominator (since h ≠ 0):
(f(a + h) - f(a))/h = 4a + 2h
Therefore, the difference quotient for f(x) = 2x² + 5 is 4a + 2h. This linear expression in a and h represents the average rate of change of the function over the interval from a to a + h. Understanding this result is crucial for grasping the concept of the derivative, which is obtained by considering what happens to this expression as h approaches 0.
In this detailed exploration, we successfully found f(a), f(a + h), and the difference quotient (f(a + h) - f(a))/h for the function f(x) = 2x² + 5. Evaluating f(a) simply involved substituting a for x, resulting in 2a² + 5. Evaluating f(a + h) required substituting (a + h) for x, expanding the squared term, and simplifying to obtain 2a² + 4ah + 2h² + 5. The most significant part was calculating the difference quotient, which we found to be 4a + 2h. This expression represents the average rate of change of the function over the interval from a to (a + h) and is a fundamental concept in calculus. This exercise has not only solidified our understanding of function evaluation but has also provided a crucial stepping stone towards understanding derivatives and the broader concepts of differential calculus. The ability to compute and interpret difference quotients is an essential skill for anyone studying calculus and related fields.
