A function f is defined on the interval [0,4], and its derivative is e^sinx-2cos3x
a. on what interval is f increasing?
b. at what value(s) of x does f have a local maxima?
c. how many points of inflection does f have?
*calculator is allowed for this problem
a. The function f is increasing on intervals where its derivative is positive. To determine these intervals, we need to find where e^sinx - 2cos3x > 0.
Let's break it down. e^sinx is always positive since e to any power is always positive. Now let's focus on -2cos3x. The range of cosine function is [-1,1], so -2cos3x lies in the range of [-2,2]. Since this is a subtraction, we are looking for x values where e^sinx is greater than 2.
To find these x values, you can use a calculator to find the inverse of e^x: ln(x), which gives us ln(2). The natural logarithm of 2 is approximately 0.693.
So, f is increasing on the interval [0, 0.693).
To find the intervals where the function is increasing, we need to find when the derivative is positive.
a. The derivative of f(x) is e^sin(x) - 2cos(3x). In order to determine when the derivative is positive, we need to find the values of x for which e^sin(x) - 2cos(3x) > 0.
Let's solve this inequality step-by-step:
e^sin(x) - 2cos(3x) > 0
We'll start by solving e^sin(x) > 2cos(3x).
Now, let's solve e^sin(x) = 2cos(3x):
Take the natural logarithm of both sides to get:
sin(x) = ln(2cos(3x))
To simplify further, let's use the identities:
cos(3x) = 4cos^3(x) - 3cos(x)
ln(2) + ln(cos(3x)) = ln(2cos(3x))
ln(2) + ln(4cos^3(x) - 3cos(x)) = ln(2cos(3x))
Now, substitute ln(2cos(3x)) with ln(4cos^3(x) - 3cos(x)):
ln(2) + ln(4cos^3(x) - 3cos(x)) = ln(4cos^3(x) - 3cos(x))
So, we have:
ln(2) = 0
Now, let's solve for x by using the inverse trigonometric functions:
sin(x) = ln(2) / ln(4cos^3(x) - 3cos(x))
Take the inverse sine of both sides:
x = sin^(-1)(ln(2) / ln(4cos^3(x) - 3cos(x)))
To find the intervals where the derivative is positive, we can use a graphing calculator or software to plot the function e^sin(x) - 2cos(3x) and observe where it crosses the x-axis.
b. To find the local maxima of f(x), we need to find the critical points where the derivative changes from positive to negative. This occurs when the derivative is equal to zero and changes sign. So, we need to find the values of x for which e^sin(x) - 2cos(3x) = 0.
By setting e^sin(x) - 2cos(3x) = 0, you can solve for x to find the values where f(x) has local maxima.
c. To determine the number of points of inflection of f(x), we need to find the values of x where the concavity of the function changes. This occurs when the second derivative changes sign.
To find the second derivative, differentiate the derivative of f(x):
f'(x) = e^sin(x) - 2cos(3x)
Take the derivative of f'(x) with respect to x:
f''(x) = e^sin(x)cos(x) + 6sin(3x)
Set f''(x) = 0 and solve for x to find the points of inflection. The number of distinct solutions will give you the number of points of inflection.
Remember, using a graphing calculator or software can help you visualize the function and better understand the nature of the intervals, local maxima, and points of inflection.
To determine the interval on which the function f is increasing, we need to analyze the sign of its derivative. Recall that if the derivative is positive, the function is increasing, and if the derivative is negative, the function is decreasing.
a. To find the interval on which f is increasing, we need to consider where the derivative is positive: e^sin(x) - 2cos(3x) > 0. Since a calculator is allowed, we can use it to solve this inequality. Here's how:
- Enter the function e^sin(x) - 2cos(3x) into your calculator.
- If your calculator supports it, plot the graph of this function.
- Analyze the graph and identify the x-values where the function is positive (above the x-axis).
- The interval(s) where the function is positive (increasing) represents where f is increasing.
b. To find the local maximum of f, we need to consider where its derivative changes sign from positive to negative (or simply becomes negative).
- On your calculator, plot the graph of the derivative function e^sin(x) - 2cos(3x).
- Analyze the graph and identify the x-values where the derivative changes sign from positive to negative (or becomes negative).
- These x-values represent the potential locations of local maxima for f.
c. To find the points of inflection of f, we need to identify where the second derivative changes sign. However, we are only given the first derivative, so we need to find the second derivative first.
- Differentiate the given derivative function, e^sin(x) - 2cos(3x), with respect to x. The result will be the second derivative.
- Once you have the second derivative, you can use your calculator to plot it and analyze where it changes sign (from positive to negative or from negative to positive).
- The x-values where the second derivative changes sign represent the potential points of inflection for f.
Remember to use your calculator to help visualize and analyze the graphs of the functions to determine the intervals, local maxima, and points of inflection.
Note: The specific steps to perform these calculations may vary depending on the calculator you are using.
a. f is increasing where f' > 0
So, where do you have e^sinx-2cos3x > 0 ?
b. you need e^sinx-2cos3x = 0
c. points of inflection are where f" = 0
f" = (e^sinx)(cosx) + 6sin3x
so, where is that zero?
you can check your answers at wolframalpha.com
The graph of e^sinx-2cos3x on [0,4] can be seen at
https://www.wolframalpha.com/input/?i=e%5Esinx-2cos3x+where+0+%3C%3D+x+%3C%3D+4