Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 25
a. Write the derivative formula. b. Locate any relative extreme points and identify the extreme as a maximum or minimum. $$ f(x)=12\left(1.5^{x}\right)+12\left(0.5^{x}\right) $$
Short Answer
Expert verified
Differentiate and solve \( f'(x) = 0 \) for critical points. Use the second derivative to classify them.
Step by step solution
01
Differentiate the Function
To find the derivative, we apply the power rule and chain rule of differentiation. The derivative of \( f(x) = 12(1.5^x) + 12(0.5^x) \) is:\[ f'(x) = 12 \cdot \frac{d}{dx} \left(1.5^x\right) + 12 \cdot \frac{d}{dx} \left(0.5^x\right) \]Using the rule \( \frac{d}{dx} (a^x) = a^x \ln(a) \), we get:\[ f'(x) = 12 \cdot (1.5^x \ln(1.5)) + 12 \cdot (0.5^x \ln(0.5)) \]
02
Set the Derivative to Zero to Find Critical Points
To locate relative extremes, set the derivative equal to zero:\[ 12 (1.5^x \ln(1.5)) + 12 (0.5^x \ln(0.5)) = 0 \]Simplify and solve:\[ 1.5^x \ln(1.5) + 0.5^x \ln(0.5) = 0 \]The critical point occurs where \( 1.5^x \ln(1.5) = -0.5^x \ln(0.5) \).
03
Solve for the Critical Point
To simplify, divide both sides by \( 1.5^x \) and rearrange:\[ \ln(1.5) = -\left( \frac{0.5}{1.5} \right)^x \ln(0.5) \]Solve for \( x \):\[ \ln(1.5) / \ln(0.5) = - \left( \frac{1}{3} \right)^x \]This equation is complex and may require numerical methods for an exact solution.
04
Determine the Nature of the Critical Point
Use the second derivative test to classify the critical point. Calculate the second derivative:\[ f''(x) = 12 \left((1.5^x \ln^2(1.5))\right) + 12 \left((0.5^x \ln^2(0.5))\right) \]Evaluate \( f''(x) \) at the critical point. If \( f''(x) > 0 \), it's a minimum; if \( f''(x) < 0 \), it's a maximum. If \( f''(x) = 0 \), the test is inconclusive. This direct calculation provides insight into the concavity at these critical points.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In calculus, the derivative represents the rate at which a function is changing at any given point. For a function like \( f(x) = 12(1.5^x) + 12(0.5^x) \), the derivative provides us insight into how the function's value changes with a slight change in \( x \). Differentiation, the process of finding the derivative, uses several rules, such as the power rule and chain rule, to compute this rate of change.
We follow these principles when differentiating exponential functions like this one. The derivative of \( a^x \) is \( a^x \ln(a) \).
This formula helps us calculate the rate of growth or decay in the function by measuring how quickly it increases or decreases. Applying these rules, we derive \( f'(x) \), which will show the function's slope at any \( x \). A positive slope (\( f'(x) > 0 \)) means the function is increasing, while a negative slope (\( f'(x) < 0 \)) indicates a decreasing function.
We follow these principles when differentiating exponential functions like this one. The derivative of \( a^x \) is \( a^x \ln(a) \).
This formula helps us calculate the rate of growth or decay in the function by measuring how quickly it increases or decreases. Applying these rules, we derive \( f'(x) \), which will show the function's slope at any \( x \). A positive slope (\( f'(x) > 0 \)) means the function is increasing, while a negative slope (\( f'(x) < 0 \)) indicates a decreasing function.
Extreme Points
Extreme points, such as maxima and minima, are crucial in calculus for understanding the behavior of functions. These points indicate where a function reaches its highest or lowest values in a given interval and can be relative or global. To find these points, we first compute the derivative and identify where it equals zero or does not exist; these are the critical points.
In the exercise, setting \( f'(x) = 0 \) allows us to find the critical points, which could be potential candidates for extreme values. To determine if these are maxima or minima, we can use the second derivative test. A positive second derivative at the critical point suggests the point is a minimum (concave up), while a negative second derivative suggests a maximum (concave down).
Understanding extreme points helps analyze not just theoretical behavior, but practical applications, such as determining profit-maximizing prices or minimizing costs in real-world problems.
In the exercise, setting \( f'(x) = 0 \) allows us to find the critical points, which could be potential candidates for extreme values. To determine if these are maxima or minima, we can use the second derivative test. A positive second derivative at the critical point suggests the point is a minimum (concave up), while a negative second derivative suggests a maximum (concave down).
Understanding extreme points helps analyze not just theoretical behavior, but practical applications, such as determining profit-maximizing prices or minimizing costs in real-world problems.
Chain Rule
The chain rule is a fundamental rule in calculus used to differentiate composite functions. When a function is composed of an outer function and an inner function, the chain rule helps us find the derivative by differentiating the outer function with respect to the inner function, and then multiplying by the derivative of the inner function.
For instance, in the function \( f(x) = 12(1.5^x) + 12(0.5^x) \), each term is an exponential function where the power \( x \) acts as an inner function. The derivative of the power involves applying the chain rule and differentiating with respect to \( x \).
This is essential when dealing with more complex expressions, especially if transformations are applied inside a function, ensuring precise calculation of derivatives. Correct application of the chain rule is vital for accurate results across various calculus problems.
For instance, in the function \( f(x) = 12(1.5^x) + 12(0.5^x) \), each term is an exponential function where the power \( x \) acts as an inner function. The derivative of the power involves applying the chain rule and differentiating with respect to \( x \).
This is essential when dealing with more complex expressions, especially if transformations are applied inside a function, ensuring precise calculation of derivatives. Correct application of the chain rule is vital for accurate results across various calculus problems.
Power Rule
The power rule is one of the simplest and most frequently used rules for differentiation. It states that if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \). Although very straightforward, it is technically applied not directly here given the nature of the function, which includes exponential terms.
In our context, the power rule is part of the broader set of rules, like exponentiation differentiation, which follows a similar logic. For exponential functions, the rule slightly adjusts, as indicated by \( \frac{d}{dx}(a^x) = a^x \ln(a) \).
Mastering the power rule, along with special cases like exponentiation, ensures a solid base for understanding how functions change, laying the groundwork for more advanced calculus concepts.
In our context, the power rule is part of the broader set of rules, like exponentiation differentiation, which follows a similar logic. For exponential functions, the rule slightly adjusts, as indicated by \( \frac{d}{dx}(a^x) = a^x \ln(a) \).
Mastering the power rule, along with special cases like exponentiation, ensures a solid base for understanding how functions change, laying the groundwork for more advanced calculus concepts.
