Problem 8
Question
Sketch a complete graph of the function. $$g(x)=(1.2)^{x}+(.8)^{-x}$$
Step-by-Step Solution
Verified Answer
Based on the previous analysis, we can conclude the following about the graph of the function $$g(x) = (1.2)^x + (0.8)^{-x}$$:
1. The domain of the function is $$(-\infty, \infty)$$, and its range is $$(0, \infty)$$.
2. The function is increasing when $$g'(x)$$ is positive and decreasing when $$g'(x)$$ is negative.
3. There is a critical point where $$g'(x) = 0$$, indicating a maximum or minimum value.
4. As $$x \rightarrow -\infty$$ and $$x \rightarrow \infty$$, $$g(x) \rightarrow \infty$$.
To sketch the graph, plot the critical point found in step 3, observe its maximum or minimum behavior, and note the function's increasing and decreasing intervals. The graph will have an increasing behavior for large positive values of $$x$$ and a decreasing behavior for large negative values of $$x$$. The transition between the two behaviors will be indicated by the critical point identified in Step 3.
1Step 1: Determine the domain and range of the function
The domain of the function $$g(x)$$ consists of all possible values of $$x$$ for which the function is defined. Since the base is positive and not equal to 1 in both exponential functions, the function is defined for all real values of $$x$$.
The range of the function is all possible values of $$g(x)$$ that the function can take. Since both exponential parts of the function are always positive and there's no restriction for $$x$$, the range of the function will be all positive real numbers.
Domain: $$(-\infty, \infty)$$
Range: $$(0, \infty)$$
2Step 2: Determine if the function is increasing or decreasing
To find the behavior of the function, we'll take the derivative of the given function and see the sign of the derivative.
The function is $$g(x) = (1.2)^x + (0.8)^{-x}$$.
Derivative of the function: $$g'(x) = (1.2)^x \cdot \ln(1.2) - (0.8)^{-x} \cdot \ln(0.8)$$
Notice that as the bases are greater than 0 and the signs of the bases are opposite, the behavior of both parts will not contradict. $$g'(x)$$ will be positive when increasing and negative when decreasing.
3Step 3: Find any critical points
To find any critical points (like maximum and minimum) or the point where the derivative changes the sign, set $$g'(x)$$ equal to zero and solve for $$x$$.
$$(1.2)^x \cdot \ln(1.2) - (0.8)^{-x} \cdot \ln(0.8) = 0$$
This equation is difficult to solve analytically. However, we can plot the derivative and visually identify its root to find the critical point.
4Step 4: Identify asymptotes or function's behavior at extremes
Analyze the function's behavior as $$x$$ approaches positive or negative infinity:
As $$x \rightarrow -\infty$$, $$g(x) \rightarrow \infty$$ because the first term goes to 0 and the second term grows uncontrollably due to the negative power.
As $$x \rightarrow \infty$$, $$g(x) \rightarrow \infty$$ because the first term grows uncontrollably, and the second term goes to 0.
5Step 5: Sketch the graph using the gathered information
Now, use the information obtained in previous steps to sketch the graph of the given function:
1. Plot the critical point found in step 3 and observe its behavior (maximum or minimum).
2. The graph will be increasing when the derivative is positive and decreasing when the derivative is negative.
3. Keep in mind the range of the function, which is \((0, \infty)\).
4. Remember the function's behavior as $$x$$ approaches infinity or negative infinity.
Based on the gathered information, sketch the curve of the function. The graph should have an increasing behavior as $$x$$ gets bigger and a decreasing behavior as $$x$$ gets smaller. The transition between the two behaviors will be indicated by the critical point identified in Step 3.
Key Concepts
DerivativeDomain and RangeGraph SketchingAsymptotes
Derivative
In calculus, the derivative of a function provides us insight into how the function behaves at every point. Specifically, it indicates the rate of change of the function’s output with respect to its input. To find the derivative of our function, which is composed of exponential terms, we utilize the basic derivative rules for exponential functions. For an exponential function like \(a^{x}\), the derivative is \(a^{x} \cdot \ln(a)\).
For the function \(g(x) = (1.2)^x + (0.8)^{-x}\), the derivative is calculated as follows:
For the function \(g(x) = (1.2)^x + (0.8)^{-x}\), the derivative is calculated as follows:
- For \((1.2)^x\), the derivative is \((1.2)^x \cdot \ln(1.2)\).
- For \((0.8)^{-x}\), the derivative is \(-(0.8)^{-x} \cdot \ln(0.8)\) due to the chain rule, accounting for the negative exponent.
Domain and Range
The domain and range of a function describe where the function is defined and what outputs it can produce. The domain of the function \(g(x) = (1.2)^x + (0.8)^{-x}\) is concerned with the values \(x\) can take so that \(g(x)\) is valid. Since we are dealing with exponential functions, the domain encompasses all real numbers, represented as \((-\infty, \infty)\).
On the other hand, the range is about the possible output values, \(g(x)\) can achieve. Each term in the function \((1.2)^x\) and \((0.8)^{-x}\) produces a positive real number, meaning the sum \(g(x)\) results in positive values as well. Thus, the range is \((0, \infty)\). This means \(g(x)\) never crosses zero, always producing positive outputs.
On the other hand, the range is about the possible output values, \(g(x)\) can achieve. Each term in the function \((1.2)^x\) and \((0.8)^{-x}\) produces a positive real number, meaning the sum \(g(x)\) results in positive values as well. Thus, the range is \((0, \infty)\). This means \(g(x)\) never crosses zero, always producing positive outputs.
Graph Sketching
When sketching the graph of an exponential function like \(g(x) = (1.2)^x + (0.8)^{-x}\), one should first consider domain, range, derivative (for determining increase/decrease behavior), and critical points.
Here’s how to approach sketching:
Here’s how to approach sketching:
- Begin with plotting known points such as the y-intercept, found by evaluating \(g(0)\).
- Observe where the derivative \(g'(x)\) changes from positive to negative or vice versa for critical points, helping identify local maxima or minima.
- Given the derivative analysis, note each section where the function is increasing or decreasing.
- Factor in asymptotic behavior at infinity as determined by earlier analysis.
Asymptotes
Asymptotes are lines that the graph of a function approaches but never touches. For our function \(g(x) = (1.2)^x + (0.8)^{-x}\), asymptotic behavior becomes evident when examining the function's limits at infinity.
As \(x\) approaches \(-\infty\), the term \((1.2)^x\) diminishes towards zero, while \((0.8)^{-x}\) grows without bound due to the negative exponent. This results in the overall function \(g(x)\) tending towards infinity. Conversely, as \(x\) approaches \(\infty\), \((1.2)^x\) grows immensely, and \((0.8)^{-x}\) nears zero, causing \(g(x)\) to again shoot towards infinity.
Thus, there are no horizontal asymptotes because \(g(x)\) isn't settling into a finite horizontal line at extremes; both ends are towards infinity. Recognizing this helps in understanding how the function behaves well beyond typical viewing spots on a graph.
As \(x\) approaches \(-\infty\), the term \((1.2)^x\) diminishes towards zero, while \((0.8)^{-x}\) grows without bound due to the negative exponent. This results in the overall function \(g(x)\) tending towards infinity. Conversely, as \(x\) approaches \(\infty\), \((1.2)^x\) grows immensely, and \((0.8)^{-x}\) nears zero, causing \(g(x)\) to again shoot towards infinity.
Thus, there are no horizontal asymptotes because \(g(x)\) isn't settling into a finite horizontal line at extremes; both ends are towards infinity. Recognizing this helps in understanding how the function behaves well beyond typical viewing spots on a graph.
Other exercises in this chapter
Problem 8
Solve the equation without using logarithms. $$5^{2 x^{2}+3 x}=25^{6-x}$$
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Translate the given logarithmic statement into an equivalent exponential statement. $$\log (.8)=-.097$$
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Simplify the expression. Assume \(a, b, c, d>0\) $$\left(a^{x^{2}}\right)^{1 / x}$$
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Solve the equation. First express your answer in terms of natural logarithms (for instance, \(x=(2+\ln 5) /(\ln 3)) .\) Then use a calculator to find an approxi
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