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Chapter 3: Problem 17

Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=e^{x}, c=0, x=-0.17 $$

Short Answer

Expert verified
The estimated value of \(f(x)\) at \(x = -0.17\) is approximately 0.83.

Step by step solution

01

Identify known values

We are given that the function is \(f(x) = e^x\) and we need to estimate \(f(x)\) at \(x = -0.17\). We know the value at \(c = 0\), which is \(f(0) = e^0 = 1\).
02

Calculate the increment

The increment \(\Delta x\) is the change from \(c\) to \(x\). Calculate this as \(\Delta x = x - c = -0.17 - 0 = -0.17\).
03

Determine the derivative at c

The derivative of \(f(x) = e^x\) is \(f'(x) = e^x\). At \(c = 0\), this gives \(f'(0) = e^0 = 1\).
04

Use the linear approximation formula

According to the method of increments, we can approximate \(f(x)\) using the formula: \(f(x) \approx f(c) + f'(c)\Delta x\). Substitute the known values: \(f(x) \approx 1 + 1 \times (-0.17)\).
05

Perform the calculation

Calculate the expression from Step 4: \(f(x) \approx 1 - 0.17 = 0.83\).

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Key Concepts

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

Linear Approximation
Linear approximation is a handy technique to estimate the value of a function at a certain point using its tangent line. Imagine you want to know how a curve behaves around a small area without going through complicated calculations. Linear approximation simplifies this by assuming the curve is nearly a straight line in that small region. This can make your calculations much quicker and easier!

Here's how it works intuitively:
  • When you have a function, say \( f(x) \), you select a point \( c \) that is close to your point of interest.
  • Then you calculate the function's value \( f(c) \) and its derivative \( f'(c) \) at that point.
  • The formula for linear approximation is:
    \[ f(x) \approx f(c) + f'(c) \Delta x \]
    where \( \Delta x = x - c \).
The idea is that the additional value \( f'(c) \Delta x \) represents the change in the function value due to moving from \( c \) to \( x \). For small changes, this provides a close estimate.
Derivative
The derivative of a function provides a measure of how the function鈥檚 output value changes as its input changes, essentially reflecting its "slope." Consider it as the function's sensitivity to changes in its variable. If you have a curve, the derivative at a point tells you how steep the curve is at that location.

Key aspects of understanding derivatives:
  • Think of the derivative \( f'(x) \) as the rate of change of \( f(x) \). It is like knowing how fast a car is moving by looking at the speedometer.
  • If \( f(x) \) is a linear function, its rate of change is constant. For curves like an exponential function, the rate can change based on \( x \).
  • To find the derivative, you use differentiation rules such as the power rule, product rule, or chain rule, depending on the function type.
In our exercise, the exponential function \( f(x) = e^x \) has the unique property where its derivative \( f'(x) = e^x \) is identical to the function itself, illustrating exponential growth.
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent, such as \( e^x \). These functions exhibit rapid growth and are used extensively in natural phenomena modeling, like population growth and radioactive decay.

Here鈥檚 why exponential functions are so intriguing:
  • The base \( e \) is a special number, approximately 2.718, known as Euler's number. It frequently appears in calculus and complex system studies.
  • An exponential function like \( e^x \) grows faster as \( x \) increases, which reflects in its steep curve on a graph.
  • One interesting property is that the derivative of \( e^x \) is itself \( e^x \), which makes it stand out among other functions.
In this exercise, the exponential function allows us to apply both the method of increments and linear approximation because of its predictable, continuous change, making it easier to estimate values close to known points.

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Most popular questions from this chapter

Differentiate the given expression with respect to \(x\). $$ \sinh ^{-1}\left(\sqrt{x^{2}-1}\right) $$

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A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x)=x^{1 / 3}, c=27, \Delta x=0.9, N=3 $$
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