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Comparing \(\Delta y\) and \(d y\) In Exercises \(7-10\) , use the information to evaluate and compare \(\Delta y\) and \(d y .\) $$ \begin{array}{ll}{\text { Function }} & {x \text { -Value }} \\ {y=x^{4}+1} & {x=-1}\end{array} \quad \Delta x=d x=0.01 $$

Short Answer

Expert verified
The results are: \(\Delta y = 0.9801\) and \(dy = -0.04\). It reveals that for small changes in \(x\) (\(dx = 0.01\)), the change in \(y\) (\(\Delta y\)) is much larger than the differential \(dy\), which is the tangent line's slope at that point.

Step by step solution

01

Find the original function value.

Start by finding the value of the original function at given \(x\). The function is \(f(x) = x^{4} + 1\). Put \(x = -1\) in \(f(x)\), which gives \(f(-1) = (-1)^{4} +1 = 1 + 1 = 2.\)
02

Compute \(\Delta y\).

Next, use the function to compute \(\Delta y = f(x + \Delta x) - f(x)\). With \(f(x) = 2\) from last step, \(x = -1\), \(\Delta x = 0.01\), we get \(\Delta y = f(-1 + 0.01) - 2 = (-0.99)^{4} + 1 - 2 = 0.9801\).
03

Compute the derivative.

To get \(dy\), first determine the derivative of the function \(f'(x) = 4x^3\). At \(x = -1\), \(f'(-1) = -4\).
04

Compute \(dy\)

Finally, use the derivative to obtain \(dy = f'(x) \cdot dx = -4 \cdot 0.01 = -0.04.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
In differential calculus, a derivative represents the rate of change of a function as its input changes. It essentially tells us how a small change in the input variable affects the output of the function. When we talk about taking a derivative, we mean finding this instantaneous rate of change.
In our exercise problem, the function given is \(y = x^4 + 1\). To find its derivative, we use the power rule. The power rule states that if \(y = x^n\), then the derivative, \(y'\) or \(f'(x)\), is \(nx^{n-1}\). For the function \(y = x^4\), \(f'(x) = 4x^3\).
Derivatives are crucial in comparing small changes, like \(dy\), allowing us to approximate how functions behave over small intervals. This makes derivatives valuable in engineering, physics, and economics, as they help understand trends and growth rates.
Function Evaluation
Evaluating a function means finding the output for a given input value. It's a crucial step in connecting calculus concepts to real numbers.
In the exercise, where \(y = x^4 + 1\), we start by evaluating this function at \(x = -1\). Plugging in the value, we calculate \(y = (-1)^4 + 1 = 2\). This gives the initial value used in the subsequent computation of changes \(\Delta y\) and \(dy\).
Function evaluation provides a concrete reference point, helping us understand how the function behaves at specific values. This methodical approach lays the groundwork for deeper comparative analysis in differential calculus.
Differential Approximation
Differential approximation allows us to estimate small changes in functions by using derivatives. The idea is to calculate the change in the output \(dy\), which is an approximate value obtained using the derivative.
In the given problem, we use the derivative \(f'(x) = 4x^3\) to find \(dy\). When \(x = -1\), the derivative \(f'(-1) = -4\). By multiplying the derivative by the small change in \(x\), which is \(dx = 0.01\), we find \(dy = -4 \times 0.01 = -0.04\).
This approximation is often more manageable and quicker than computing actual changes, especially for complex functions. It's a practical tool when the change in \(x\) is tiny, leading to simpler solutions and insights in calculus.
Delta Notation
Delta notation is used to signify change or difference in mathematics. Specifically, \(\Delta y\) represents the actual change in the value of the function when \(x\) changes by \(\Delta x\). It measures the difference between new and original function values.
For the function \(f(x) = x^4 + 1\), when \(x = -1\) and \(\Delta x = 0.01\), calculating \(\Delta y\) involves evaluating the function at \(x + \Delta x = -0.99\). The change can be expressed as \(\Delta y = f(-0.99) - f(-1) = (-0.99)^4 + 1 - 2\). This gives \(\Delta y = 0.9801\).
Delta notation is intuitive for representing exact changes and is used alongside differentials to compare precise and approximate changes efficiently. It holds significance in mathematical modeling, where precision is crucial in simulations and predictions.

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Most popular questions from this chapter

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Comparing \(\Delta y\) and \(d y\) In Exercises \(7-10\) , use the information to evaluate and compare \(\Delta y\) and \(d y .\) Comparing \(\Delta y\) and \(d y\) In Exercises \(7-10\) , use the information to evaluate and compare \(\Delta y\) and \(d y .\) $$ \begin{array}{ll}{\text { Function }} & {x \text { -Value }} \\ {y=x^{3}} & {x=1}\end{array} \quad \Delta x=d x=0.1 $$

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