Problem 11
Question
Evaluate each function at the given point. \(h\left(x_{1}, x_{2}\right)=x_{2} e^{-x_{1} / x_{2}}\) at \((2,-1)\)
Step-by-Step Solution
Verified Answer
\(h(2, -1) = - e^{2}\) or approximately \(-7.389\).
1Step 1: Identify the function and the given point
The function is given as \(h(x_1, x_2) = x_2 e^{-x_1 / x_2}\). The point at which we need to evaluate the function is \((2, -1)\). Here, \(x_1 = 2\) and \(x_2 = -1\).
2Step 2: Substitute the values into the function
Substitute \(x_1 = 2\) and \(x_2 = -1\) into the function \(h(x_1, x_2)\). This gives us:\[ h(2, -1) = (-1) e^{-2 / (-1)} \]
3Step 3: Simplify the exponent
Simplify the exponent by dividing \(-2\) by \(-1\):\[ e^{-2 / (-1)} = e^{2} \]
4Step 4: Compute the value using simplification
Now substitute back into the function:\[ h(2, -1) = (-1) e^{2} \]This simplifies to:\[ h(2, -1) = - e^{2} \]
5Step 5: Calculate the numerical value (optional)
If needed, use a calculator to find the numerical value of \(- e^{2}\). The value of \(e\) is approximately 2.718, so:\[ h(2, -1) \approx - (2.718)^2 = -7.389 \]
Key Concepts
Exponential FunctionFunction EvaluationCoordinate Substitution
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. It can be recognized in forms like \(a^x\), where \(a\) is a positive constant and \(x\) is a variable. In our original exercise, the function is \(x_2 e^{-x_1 / x_2}\). Here, \(e\) is the base of the natural logarithm, and it is an important constant in mathematics, approximately equal to 2.718.
Exponential functions are unique because they have applications across various fields such as biology, finance, and physics due to their characteristic growth or decay rates. For example, in the provided exercise, \(e^{-x_1 / x_2}\) demonstrates exponential behavior because it involves \(e\) raised to a power determined by \(-x_1 / x_2\). This structure allows the function to model real-world scenarios where changes occur at an exponential rate.
Understanding exponential functions involves grasping how the exponent impacts the expression's value. A negative exponent, as seen in the exercise, indicates rapid growth if we simplify \(e^{-x_1 / x_2}\) to \(e^2\) upon substitution. This part of the function affects how the output behaves dynamically based on the inputs \(x_1\) and \(x_2\).
Exponential functions are unique because they have applications across various fields such as biology, finance, and physics due to their characteristic growth or decay rates. For example, in the provided exercise, \(e^{-x_1 / x_2}\) demonstrates exponential behavior because it involves \(e\) raised to a power determined by \(-x_1 / x_2\). This structure allows the function to model real-world scenarios where changes occur at an exponential rate.
Understanding exponential functions involves grasping how the exponent impacts the expression's value. A negative exponent, as seen in the exercise, indicates rapid growth if we simplify \(e^{-x_1 / x_2}\) to \(e^2\) upon substitution. This part of the function affects how the output behaves dynamically based on the inputs \(x_1\) and \(x_2\).
Function Evaluation
Function evaluation refers to the process of calculating the output of a function for given inputs. It involves substituting the specified values directly into the function's formula. In the exercise, we evaluate \(h(x_1, x_2) = x_2 e^{-x_1 / x_2}\) at the point \((2, -1)\).
The first step in function evaluation is to clearly identify and substitute the given variables into the function. Here, \(x_1 = 2\) and \(x_2 = -1\), which results in the transformation of \(h(x_1, x_2)\) to \((-1) e^{-2 / (-1)}\). By simplifying this, we substitute back to form \(-e^2\) as a cleaner expression.
The key to mastering function evaluation is breaking it down into systematic steps:
The first step in function evaluation is to clearly identify and substitute the given variables into the function. Here, \(x_1 = 2\) and \(x_2 = -1\), which results in the transformation of \(h(x_1, x_2)\) to \((-1) e^{-2 / (-1)}\). By simplifying this, we substitute back to form \(-e^2\) as a cleaner expression.
The key to mastering function evaluation is breaking it down into systematic steps:
- Identify the formula and given inputs.
- Substitute these values precisely.
- Simplify step-by-step to avoid errors.
Coordinate Substitution
Coordinate substitution is a technique used in multivariable calculus to replace variables within a function with specific values defined by a coordinate point. It effectively simplifies the problem by turning a general function into a specific numerical expression.
In the case of the given exercise, we begin with the function \(h(x_1, x_2) = x_2 e^{-x_1 / x_2}\) and are required to evaluate it at the point \((2, -1)\). This involves substituting \(x_1\) with 2 and \(x_2\) with -1. The substitution transforms the function into \((-1) e^{2}\).
The use of coordinate substitution is crucial in exploring the behavior of functions across different inputs:
In the case of the given exercise, we begin with the function \(h(x_1, x_2) = x_2 e^{-x_1 / x_2}\) and are required to evaluate it at the point \((2, -1)\). This involves substituting \(x_1\) with 2 and \(x_2\) with -1. The substitution transforms the function into \((-1) e^{2}\).
The use of coordinate substitution is crucial in exploring the behavior of functions across different inputs:
- It provides a tangible numeric output, helping grasp the function's behavior at that point.
- It challenges understanding by simplifying complex multivariable expressions into more manageable forms.
Other exercises in this chapter
Problem 11
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