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In Exercises \(15-22,\) use analytic methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. $$f(x)=e^{2 x}$$

Short Answer

Expert verified
The function \(f(x)=e^{2x}\) doesn't have any local extrema. It is increasing for all real numbers and it is not decreasing in any interval.

Step by step solution

01

Find the Derivative

The first step is to compute the derivative of the given function \(f(x)=e^{2x}\). Using the chain rule of derivatives \( \frac{d}{dx}e^{u} = e^{u} \cdot \frac{du}{dx}\), the derivative \(f'(x)\) of \(f(x) = e^{2x}\), will be \(f'(x) = 2e^{2x}\)
02

Find the Critical Points

Next, obtain the critical points. These are the points where the derivative is zero or undefined. However, for our case, the derivative \(f'(x)=2e^{2x}\) is always positive for all real values of x, and it's not undefined at any point. So, there are no critical points.
03

Determine whether the function is increasing or decreasing

The function is increasing if the derivative is positive and decreasing if it's negative. Since the derivative \(f'(x)=2e^{2x}\) is always positive for all real values of x, the function is increasing for all real numbers. Therefore, the function does not have any local extrema nor any intervals where it is decreasing.

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus. It represents the rate at which a function changes. For the function \( f(x) = e^{2x} \), to find its derivative, we use the chain rule. The chain rule states that the derivative of \( e^u \) with respect to \( x \) is \( e^u \cdot \frac{du}{dx} \). For our function, \( u = 2x \), so \( \frac{du}{dx} = 2 \). This gives us the derivative of \( f(x) \) as \( f'(x) = 2e^{2x} \).

Here are the essential steps in deriving this function:
  • Identify the inner function: In \( e^{2x} \), the inner function is \( 2x \).
  • Differentiate the inner function: \( \frac{d}{dx} (2x) = 2 \).
  • Multiply by the derivative of the inner function: \( e^{2x} \cdot 2 \).

The derivative signifies the slope of the tangent line to the curve at any given point along the curve. For our function, knowing that the derivative \( 2e^{2x} \) is always positive means that the function is increasing everywhere.
Local Extrema
Local extrema are points where a function reaches a local maximum or minimum. It's where the function "levels out" momentarily before changing direction. However, to find these points, the derivative needs to be zero or undefined at some point.

In the case of \( f(x) = e^{2x} \), the derivative \( f'(x) = 2e^{2x} \) is always positive. It never equals zero and is never undefined. Therefore, no local extrema exist for this function. This means that there are no peaks or troughs on the graph.

To summarize:
  • Local extrema occur where \( f'(x) = 0 \) or is undefined.
  • For \( e^{2x} \), \( f'(x) = 2e^{2x} \) is always positive.
  • Thus, no local maxima or minima are present in the function.
Increasing and Decreasing Intervals
Understanding where a function is increasing or decreasing helps visualize its behavior. A function is increasing if its derivative is positive, while it is decreasing if the derivative is negative.

For \( f(x) = e^{2x} \), the derivative \( f'(x) = 2e^{2x} \) is positive for all real values of \( x \). Thus, the function is increasing on the entire domain of all real numbers. This means the curve of \( f(x) \) continuously rises as \( x \) increases.

Key points to note:
  • An increasing function has \( f'(x) > 0 \).
  • A decreasing function would have \( f'(x) < 0 \). However, \( f'(x) = 2e^{2x} \) never becomes negative.
  • Therefore, \( f(x) = e^{2x} \) increases for all \( x \).

This concept helps predict the behavior of functions and their graphs, ensuring a solid understanding of calculus principles.

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