Problem 112
Question
Graph \(f, f^{\prime},\) and \(f^{\prime \prime}\) $$ f(x)=1000 e^{-0.08 x} $$
Step-by-Step Solution
Verified Answer
Graph exponential decay for all: \( f(x)=1000e^{-0.08x}, f'(x)=-80e^{-0.08x}, f''(x)=6.4e^{-0.08x} \).
1Step 1: 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 \( x \).
2Step 2: 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, \( f'(x) = 1000 \times -0.08 \times e^{-0.08x} = -80e^{-0.08x} \).
3Step 3: 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 \( f''(x) = -80 \times -0.08 \times e^{-0.08x} = 6.4e^{-0.08x} \).
4Step 4: 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.
Key Concepts
DerivativeRate of ChangeConcavityGraphing Functions
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:
For our function, which is an exponential decay function given by:
- \( f(x) = 1000 e^{-0.08x} \)
- \( f'(x) = -80e^{-0.08x} \)
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 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:
- \( f''(x) = 6.4e^{-0.08x} \)
- 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.
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:
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.
Other exercises in this chapter
Problem 111
Graph \(f, f^{\prime},\) and \(f^{\prime \prime}\) $$ f(x)=2 e^{0.3 x} $$
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Differentiate. $$ y=\ln \left|\frac{x^{5}}{(8 x+5)^{2}}\right| $$
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Graph $$ f(x)=\left(1+\frac{1}{x}\right)^{x} $$ Use the TABLE feature and very large values of \(x\) to confirm that \(e\) is approached as a limit.
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