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Chapter 3: Problem 112

Graph \(f, f^{\prime},\) and \(f^{\prime \prime}\) $$ f(x)=1000 e^{-0.08 x} $$

Short Answer

Expert verified
Graph exponential decay for all: \( f(x)=1000e^{-0.08x}, f'(x)=-80e^{-0.08x}, f''(x)=6.4e^{-0.08x} \).

Step by step solution

01

Identify the Function

The given function is the exponential function \( f(x) = 1000 e^{-0.08x} \). This function models exponential decay because it has a negative exponent associated with \( x \).
02

Compute the First Derivative

The first derivative of the function \( f(x) \) represents the rate of change of \( f \) with respect to \( x \). For an exponential function of the form \( ae^{bx} \), the derivative is \( abe^{bx} \). In this case, \( f'(x) = 1000 \times -0.08 \times e^{-0.08x} = -80e^{-0.08x} \).
03

Compute the Second Derivative

The second derivative, \( f''(x) \), gives us the rate of change of the first derivative, and it indicates the concavity of the function. Differentiating \( f'(x) = -80e^{-0.08x} \) using the same exponential rule, we get \( f''(x) = -80 \times -0.08 \times e^{-0.08x} = 6.4e^{-0.08x} \).
04

Sketch the Functions

Graphing these functions involves plotting \( f(x) = 1000e^{-0.08x} \), \( f'(x) = -80e^{-0.08x} \), and \( f''(x) = 6.4e^{-0.08x} \). Since all are exponential functions with the same exponent, they will exhibit similar shapes (decay curves), but with different initial values (1000, -80, 6.4) affecting their steepness and direction.

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Key Concepts

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

Derivative
The derivative of a function represents the rate at which a function is changing at any given point. It is like capturing the speed of a car at a precise moment while it's moving.
For our function, which is an exponential decay function given by:
  • \( f(x) = 1000 e^{-0.08x} \)
The first derivative, \( f'(x) \), tells us how quickly the function is decreasing. For exponential functions of the format \( ae^{bx} \), the derivative follows the pattern \( abe^{bx} \). In our specific case, the calculation was:
  • \( f'(x) = -80e^{-0.08x} \)
This derivative shows that the function decreases, with the negative sign indicating the direction of the change.
Rate of Change
The rate of change of a function determines how a function's output values change over its input values. It gives insight into the speed or intensity of this change. In the context of our exponential decay function:
  • \( f(x) = 1000 e^{-0.08x} \)
The rate of change is continuously decreasing because of the negative sign in the derivative \( f'(x) = -80e^{-0.08x} \). This means the intensity of the decay is constant, as the derivative is also an exponential decay function:
  • The initial rate of change is -80 (from the factor before the exponential).
  • As \( x \) increases, the values of \( e^{-0.08x} \) decrease, showing slower rates of decay over time.
Concavity
Concavity describes how the curvature of a function behaves, and the second derivative provides us this information. For the given function \( f(x) = 1000 e^{-0.08x} \), the second derivative is:
  • \( f''(x) = 6.4e^{-0.08x} \)
This positive outcome tells us about the concavity:
  • The function \( f(x) \) is concave up, meaning it bends upwards as \( x \) increases.
  • The positive value indicates a gentle decreasing curve but still, in the increasing direction for the second derivative, this encourages the overall upward bend.
Understanding concavity reveals how a function might change direction or the robustness of its decreases or increases.
Graphing Functions
Graphing functions of \( f(x) = 1000 e^{-0.08x} \), along with its derivatives \( f'(x) = -80e^{-0.08x} \) and \( f''(x) = 6.4e^{-0.08x} \), allows visual interpretation of all concepts we've discussed above.
Here's what to keep in mind when graphing these:
  • All three functions exhibit exponential decay behavior, thus they appear similar in shape.
  • The initial values of the functions cause different steepness: \( f(x) \) starts at 1000, \( f'(x) \) at -80, and \( f''(x) \) at 6.4, affecting their vertical stretch.
  • Each graph shows progressively slower declines as \( x \) increases.
Visual graphs transform these mathematical attributes into more tangible interpretations, making them easier to understand.

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Most popular questions from this chapter

Seventeen adults came ashore from the British ship HMS Bounty in 1790 to settle on the uninhabited South Pacific island Pitcairn. The population, \(P(t),\) of the island \(t\) years after 1790 can be approximated by the logistic equation $$ P(t)=\frac{3400}{17+183 e^{-0.982 t}} $$ a) Find the population of the island after \(10 \mathrm{yr}, 50 \mathrm{yr}\) and \(75 \mathrm{yr}\) b) Find the rate of change in the population, \(P^{\prime}(t)\). c) Find the rate of change in the population after \(10 \mathrm{yr}, 50 \mathrm{yr},\) and \(75 \mathrm{yr}\) d) What is the limiting value for the population of Pitcairn? (The limiting value is the number to which the population gets closer and closer but never reaches.)

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