Problem 44

Question

Find the average rate of change of each ficnetion on the given interval. $$f(x)=-2 x^{2}+5 ; \text { interval: }[-2,-1]$$

Step-by-Step Solution

Verified
Answer
The average rate of change of function f(x) over the interval [-2, -1] is 6
1Step 1: Evaluate the function at the endpoints of the interval
To find the average rate of change, we first need to evaluate the function at the endpoints of the given interval, -2 and -1. \n \[ f(-2) = -2(-2)^2 + 5 = -2*4 + 5 = -3 \] \n And, \n \[ f(-1) = -2(-1)^2 + 5 = -2*1 + 5 = 3 \]
2Step 2: Apply the formula for average rate of change
Now that we have the values of the function at the endpoints of the interval, we can calculate the average rate of change using the formula: \n \[ \frac{f(b) - f(a)}{b - a} \] \n Substituting the values we obtained in Step 1 into this formula, we get: \n \[ \frac{f(-1) - f(-2)}{-1 - (-2)} = \frac{3 - (-3)}{-1 + 2} \]
3Step 3: Compute the average rate of change
Finally we calculate the average rate of change by simplifying the expression from Step 2: \[ \frac{6}{1} = 6 \]

Key Concepts

Understanding Quadratic FunctionsEvaluating FunctionsInterval Calculations for Average Rate of Change
Understanding Quadratic Functions
A quadratic function is a polynomial function of degree two and is generally represented in the form \( f(x) = ax^2 + bx + c \). In the quadratic function provided in the exercise, we have \( f(x) = -2x^2 + 5 \). Here, the coefficient \(-2\) is associated with \(x^2\), and it determines the shape and direction of the parabola that the function graphs.
- If \(a\) (the coefficient of \(x^2\)) is positive, the parabola opens upwards; if \(a\) is negative, the parabola opens downwards. In this case, since \(a = -2\), the parabola opens downwards.
- The constant term \(c\) provides the y-intercept of the graph, which in this case is 5.
Quadratic functions have a variety of applications including physics and engineering, where they can model natural phenomena like projectile motion.Understanding the structure of a quadratic function can help with operations like finding roots, analyzing symmetry, or understanding the function's behavior over different intervals.
Evaluating Functions
Evaluating a function involves substituting the input variable \(x\) with a specific value, then calculating the output. It's like checking the functional output for given input values.
For our problem above, to evaluate at \(x = -2\), we substitute \(-2\) into the quadratic function:
  • \( f(-2) = -2(-2)^2 + 5 = -3 \)
  • \(-2^2\) results in 4, and then multiplying by \(-2\) gives \(-8\), adding 5 gives \(-3\).
Similarly, evaluating at \(x = -1\), we substitute and calculate:
  • \( f(-1) = -2(-1)^2 + 5 = 3 \)
  • \(-1^2\) results in 1, then multiplying by \(-2\) gives \(-2\), and adding 5 gives \(3\).
Evaluating functions is essential in determining behavior of functions at specific points, allowing analysis on how changes in input affect the output.
Interval Calculations for Average Rate of Change
Finding the average rate of change involves comparing how much a function's value changes between two points, divided by the interval length between these points. This provides insight into how steep a function is on that interval, resembling the concept of slope from linear algebra.
In our exercise, the formula for average rate of change is given by: \[ \frac{f(b) - f(a)}{b - a} \] where \(a\) and \(b\) are the interval's endpoints. Here, this becomes \([-2, -1]\).
  • The calculated values from function evaluation were \(f(-2) = -3\) and \(f(-1) = 3\).
  • Substituting: \(\frac{3 - (-3)}{-1 - (-2)} = \frac{6}{1}\).
  • This simplifies to 6, indicating that on average, the function increases by 6 units for every one unit it moves to the right on the interval from \(-2\) to \(-1\).
Understanding interval calculations for functions is crucial in fields like calculus, where they help predict how a function may behave in continuous change scenarios.