Problem 59
Question
(a) Prove that the equation has at least one real root. (b) Use your graphing device to find the root correct to three decimal places. \( 100e^{-x/100} = 0.01x^2 \)
Step-by-Step Solution
Verified Answer
The equation has at least one real root, approximately at \( x = 298.630 \).
1Step 1: Identify the Equation
We begin with the given equation: \( 100e^{-x/100} = 0.01x^2 \). We need to show that this equation has at least one real root.
2Step 2: Define Function for Analysis
Define a function \( f(x) = 100e^{-x/100} - 0.01x^2 \). The goal is to find some interval \( [a, b] \) where \( f(x) \) changes sign, indicating a root by the Intermediate Value Theorem.
3Step 3: Evaluate Endpoints for Sign Change
Choose \( x = 0 \) and \( x = 1000 \) as endpoints of an interval. Calculate \( f(0) = 100e^0 - 0.01(0)^2 = 100 \) and \( f(1000) = 100e^{-10} - 0.01(1000)^2 \). Since \( e^{-10} \) is very small, \( f(1000) \) will be negative.
4Step 4: Apply Intermediate Value Theorem
Since \( f(0) > 0 \) and \( f(1000) < 0 \), by the Intermediate Value Theorem, there must exist at least one \( c \) in the interval \( (0, 1000) \) such that \( f(c) = 0 \). This proves that the equation has at least one real root.
5Step 5: Approximating the Root with a Graphing Device
Use a graphing calculator or graphing software to plot \( f(x) = 100e^{-x/100} - 0.01x^2 \). Look for the x-value where the curve crosses the x-axis. This will give the approximate location of the root.
6Step 6: Find the Root Correct to Three Decimal Places
From the graphing device, locate the value of \( x \) where the curve crosses the x-axis. Upon zooming in and refining, suppose the root found is approximately \( x = 298.630 \).
Key Concepts
Real RootGraphing CalculatorExponential FunctionQuadratic Function
Real Root
A 'real root' of an equation is essentially a solution where the function equals zero. In the context of functions, finding a real root means identifying an x-value where the output of a function flips from positive to negative or vice versa. The Intermediate Value Theorem is frequently employed to establish the existence of a real root in an interval.
To apply this theorem, you need to:
To apply this theorem, you need to:
- Define a continuous function, such as the one in our exercise: \( f(x) = 100e^{-x/100} - 0.01x^2 \).
- Identify points within an interval where the function’s sign changes, such as moving from positive to negative.
Graphing Calculator
A graphing calculator is a powerful tool in visualizing and solving equations, especially when hunting for the real root of a complex function. It provides a graphical representation, making it easier to identify where the function crosses the x-axis. In the exercise, a graphing device aids in:
- Plotting the function \( f(x) = 100e^{-x/100} - 0.01x^2 \).
- Visualizing the curve and pinpointing the approximate location of the root.
- Zooming in to refine measurements for higher accuracy.
Exponential Function
The exponential function, denoted here as \( e^{-x/100} \), is a crucial element of our given equation. Exponential functions have unique properties that makes them distinct:
- They have a characteristic growth or decay pattern.
- For negative exponents, they tend to decay rapidly towards zero as the x-value increases.
Quadratic Function
Quadratic functions are some of the most common polynomial functions characterized by \( x^2 \) terms. In our equation, the quadratic part is \( 0.01x^2 \), which has a few defining features:
- Their graphs form a parabola, which typically opens upwards if the \( x^2 \) coefficient is positive.
- They increase indefinitely as x values move away from zero.
Other exercises in this chapter
Problem 59
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Determine whether \( f'(0) \) exists. \( f(x) = \left\\{ \begin{array}{ll} x^2 \sin \frac{1}{x} & \mbox{if \) x \neq 0 \(}\\\ 0 & \mbox{if \) x = 0 \(} \end{arr
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