Problem 19
Question
\(17-28\) A function is given. Determine the average rate of change of the function between the given values of the variable. $$ h(t)=t^{2}+2 t ; \quad t=-1, t=4 $$
Step-by-Step Solution
Verified Answer
The average rate of change of the function is 5.
1Step 1: Understand the Average Rate of Change Formula
The average rate of change of a function between two values \(a\) and \(b\) is given by the formula:\[\frac{f(b) - f(a)}{b - a}\]Substitute \(a = -1\) and \(b = 4\) in the function \(h(t) = t^2 + 2t\).
2Step 2: Calculate h(-1)
Substitute \(t = -1\) into the function to find \(h(-1)\):\[h(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1\]
3Step 3: Calculate h(4)
Substitute \(t = 4\) into the function to find \(h(4)\):\[h(4) = 4^2 + 2(4) = 16 + 8 = 24\]
4Step 4: Apply the Average Rate of Change Formula
Insert the calculated values \(h(-1) = -1\) and \(h(4) = 24\) into the average rate of change formula:\[\frac{h(4) - h(-1)}{4 - (-1)} = \frac{24 - (-1)}{4 + 1} = \frac{25}{5} = 5\]
Key Concepts
Quadratic FunctionsRate of Change FormulaFunction Evaluation
Quadratic Functions
Quadratic functions are an essential class of polynomial functions characterized by their highest degree term being a square, or having an exponent of two. These functions are typically written in the standard form
\( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \).
The defining feature of a quadratic function is its parabolic graph, which can open either upwards or downwards, depending on the sign of \( a \). If \( a \) is positive, the parabola opens upwards, creating a U-shape, while a negative \( a \) results in a downward-opening parabola, resembling an upside-down U.
\( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \).
The defining feature of a quadratic function is its parabolic graph, which can open either upwards or downwards, depending on the sign of \( a \). If \( a \) is positive, the parabola opens upwards, creating a U-shape, while a negative \( a \) results in a downward-opening parabola, resembling an upside-down U.
- Quadratics exhibit symmetry about a vertical axis known as the "axis of symmetry." This line passes through the vertex, the highest or lowest point on the graph.
- The vertex can be a maximum or minimum point, depending on the direction the parabola opens.
- The function's graph crosses the y-axis at the point \((0, c)\), called the y-intercept.
Rate of Change Formula
The rate of change formula is a crucial mathematical tool used to determine how a quantity changes over an interval. For functions, especially, it tells us the average change in the function's output per unit change in input. For a function defined by \( f(x) \), the average rate of change from \( x = a \) to \( x = b \) is given by
\[\frac{f(b) - f(a)}{b - a} \]This formula essentially computes the slope of the line, or "secant line," connecting the points \((a, f(a))\) and \((b, f(b))\) on the graph of the function.
\[\frac{f(b) - f(a)}{b - a} \]This formula essentially computes the slope of the line, or "secant line," connecting the points \((a, f(a))\) and \((b, f(b))\) on the graph of the function.
- If the rate is positive, the function is increasing over the interval \([a, b]\).
- If the rate is negative, the function is decreasing over that interval.
- When the rate is zero, it indicates no change over the interval.
Function Evaluation
Function evaluation is the process of calculating the output of a function given a specific input value. This process involves substituting the input value into the function's expression and performing the necessary arithmetic operations. For instance, consider the function
\( h(t) = t^2 + 2t \) from the example. To evaluate this function at \( t = -1 \), substitute -1 for \( t \):
\[h(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1\]Similarly, evaluating the same function at \( t = 4 \) involves substitution as follows:
\[h(4) = 4^2 + 2(4) = 16 + 8 = 24\]Function evaluation is crucial not only for understanding how specific inputs affect outputs but also for calculating changes over intervals.
\( h(t) = t^2 + 2t \) from the example. To evaluate this function at \( t = -1 \), substitute -1 for \( t \):
\[h(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1\]Similarly, evaluating the same function at \( t = 4 \) involves substitution as follows:
\[h(4) = 4^2 + 2(4) = 16 + 8 = 24\]Function evaluation is crucial not only for understanding how specific inputs affect outputs but also for calculating changes over intervals.
- This evaluation process is foundational in calculus, where concepts such as derivatives and integrals extend function analysis.
- Evaluating functions also aids in solving systems of equations and modeling real-world scenarios with mathematical expressions.
Other exercises in this chapter
Problem 19
Sketch the graph of the function by first making a table of values. $$ f(x)=|2 x-2| $$
View solution Problem 19
If \(f(x)=5-2 x,\) find \(f^{-1}(3)\)
View solution Problem 19
Evaluate the function at the indicated values. $$ \begin{array}{l}{f(x)=2|x-1|} \\ {f(-2), f(0), f\left(\frac{1}{2}\right), f(2), f(x+1), f\left(x^{2}+2\right)}
View solution Problem 20
\(17-22=\) Use \(f(x)=3 x-5\) and \(g(x)=2-x^{2}\) to evaluate the expression. $$ \begin{array}{ll}{\text { (a) }(f \circ f)(-1)} & {\text { (b) }(g \circ g)(2)
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