case 1. Find the absolute maxima and minina values of
f(x)=sin2x+cos2x on [0, pie] Specify both the x and y coordinateof the absolute maximum and absolute minimum. Gives answers in exact form.
case 2. Use the second derivative test for relative maxima nd minima to find the points where the function
f(x)=x^4/4-2x^2+1 has a relative maxima and relative minima. give both the x and y coordinates of the relative extrema.
case 3. Determine for which values of x in [o, pie] the tangent line to the graph of f(x)= sinx/2+cosx is horizantal. Gice answers in exact form.
All of these are straightforward. What is your question? We will be happy to critique your work or thinking. We wont do work for you. On the test questions above, number three remember the slope will be zero if horizontal. So, solve for f'=0, and remember sinx/cosx is tangent x.
I don't know how to do these, teacher is bad. Please explain to me how to do them.
Hmmm.
1. Take the derivative, set to zero. Solve for x at these points (you can compute y). Use the second derivative to see if it is a max or min.
case 2. Basically, same as case 1.
If you cannot take the derivative, or solve the f'=0 equations, that is another matter, and you need to learn how to do that pronto.
Have you looked at Schaum's Outline Series, College Calculus ? It is available at any college bookstore, or Barnes Noble. Take a look at this well done problem book. It is cheap to look, and cheap to buy.
gfind the global maximum and minimum and for the f(x)=e^X+cos on closed interval {0,pi}
gfind the global maximum and minimum and for the f(x)=e^X+cosx on the closed interval {0,pi}
Sure, I can help you understand how to solve these problems step-by-step. Let's start with Case 1.
Case 1: Find the absolute maximum and minimum values of f(x) = sin(2x) + cos(2x) on the interval [0, π].
Step 1: Take the derivative of f(x) with respect to x:
f'(x) = (d/dx)(sin(2x) + cos(2x)) = 2cos(2x) - 2sin(2x).
Step 2: Set the derivative equal to zero and solve for x to find the critical points:
2cos(2x) - 2sin(2x) = 0.
Step 3: Simplify the equation:
cos(2x) = sin(2x).
Step 4: Use the identity sin2x + cos2x = 1 to rewrite the equation:
1 = 2sin(2x)cos(2x).
Step 5: Solve for x by considering two cases:
Case 1: when sin(2x) = 0:
This implies that x = 0 or x = π/4.
Case 2: when cos(2x) = 1/2:
This implies that 2x = π/3, 5π/3, or 7π/3, so x = π/6, 5π/6, or 7π/6.
Therefore, the critical points are x = 0, π/4, π/6, 5π/6, 7π/6.
Step 6: Calculate the function values at the critical points and endpoints:
f(0) = sin(2(0)) + cos(2(0)) = 0 + 1 = 1.
f(π/4) = sin(2(π/4)) + cos(2(π/4)) = 1/√2 + 1/√2 = √2.
f(π/6) = sin(2(π/6)) + cos(2(π/6)) = 1/2 + √3/2 = (1 + √3)/2.
f(5π/6) = sin(2(5π/6)) + cos(2(5π/6)) = 1/2 - √3/2 = (1 - √3)/2.
f(7π/6) = sin(2(7π/6)) + cos(2(7π/6)) = -1/2 - √3/2 = -(1 + √3)/2.
f(π) = sin(2π) + cos(2π) = 0 + 1 = 1.
Step 7: Determine the maximum and minimum values:
The maximum value is √2, which occurs at x = π/4.
The minimum value is -(1 + √3)/2, which occurs at x = 7π/6.
So, the absolute maximum is (√2, π/4) and the absolute minimum is (-(1 + √3)/2, 7π/6).
Let's move on to Case 2.
Case 2: Use the second derivative test for relative maxima and minima to find the points where the function f(x) = x^4/4 - 2x^2 + 1 has a relative maximum and relative minimum.
Step 1: Take the derivative of f(x) with respect to x:
f'(x) = (d/dx)(x^4/4 - 2x^2 + 1) = x^3 - 4x.
Step 2: Set the derivative equal to zero and solve for x to find the critical points:
x^3 - 4x = 0.
Step 3: Factor out x:
x(x^2 - 4) = 0.
Step 4: Solve for x by considering two cases:
Case 1: when x = 0.
Case 2: when x^2 - 4 = 0. This implies that x = 2 or x = -2.
Therefore, the critical points are x = -2, 0, and 2.
Step 5: Take the second derivative of f(x):
f''(x) = (d^2/dx^2)(x^4/4 - 2x^2 + 1) = (d/dx)(x^3 - 4x) = 3x^2 - 4.
Step 6: Evaluate the second derivative at the critical points:
f''(-2) = 3(-2)^2 - 4 = 12 - 4 = 8 (positive, indicating a relative minimum).
f''(0) = 3(0)^2 - 4 = -4 (negative, indicating a relative maximum).
f''(2) = 3(2)^2 - 4 = 12 - 4 = 8 (positive, indicating a relative minimum).
Therefore, the point (2, -2) is a relative minimum and the point (0, 1) is a relative maximum.
Finally, let's tackle Case 3.
Case 3: Determine for which values of x in [0, π] the tangent line to the graph of f(x) = sin(x)/2 + cos(x) is horizontal.
Step 1: Calculate the derivative of f(x):
f'(x) = (d/dx)(sin(x)/2 + cos(x)) = (1/2)cos(x) - sin(x).
Step 2: Set the derivative equal to zero and solve for x to find where the tangent line is horizontal:
(1/2)cos(x) - sin(x) = 0.
Step 3: Multiply through by 2 to eliminate the fraction:
cos(x) - 2sin(x) = 0.
Step 4: Use the identity sin(x)/cos(x) = tan(x) to rewrite the equation:
cos(x) = 2sin(x) => cos(x) = 2(sin(x)/cos(x)) => cos^2(x) = 2sin(x).
Step 5: Use the identity cos^2(x) + sin^2(x) = 1 to substitute for cos^2(x):
1 - sin^2(x) = 2sin(x).
Step 6: Rearrange the equation and solve for sin(x):
sin^2(x) + 2sin(x) - 1 = 0.
Step 7: Solve the quadratic equation using factoring, completing the square, or the quadratic formula:
(sin(x) + 1)(sin(x) - 1) = 0.
Step 8: Solve for sin(x) by considering two cases:
Case 1: when sin(x) + 1 = 0:
This implies that sin(x) = -1, which occurs at x = 3π/2.
Case 2: when sin(x) - 1 = 0:
This implies that sin(x) = 1, which occurs at x = π/2.
Therefore, the tangent line to the graph of f(x) = sin(x)/2 + cos(x) is horizontal at x = π/2 and 3π/2.
I hope this helps you understand how to solve these problems step-by-step. Let me know if you have any further questions!
Sure, I can explain how to approach each of these problems.
Case 1: Finding the absolute maxima and minima of f(x) = sin^2(x) + cos^2(x) on [0, π]
To find the absolute maxima and minima of a function on a closed interval, you need to find the critical points and endpoints of the interval, and then evaluate the function at these points to determine the maximum and minimum values.
Here's how you can do it step by step:
1. Take the derivative of f(x): f'(x) = 2sin(x)cos(x) - 2sin(x)cos(x) = 0.
2. Solve the equation f'(x) = 0 to find the critical points:
2sin(x)cos(x) - 2sin(x)cos(x) = 0
0 = 0
Since the equation is always true, there are no critical points in the interval [0, π].
3. Evaluate the function at the endpoints of the interval:
f(0) = sin^2(0) + cos^2(0) = 0 + 1 = 1
f(π) = sin^2(π) + cos^2(π) = 0 + 1 = 1
4. Compare the values obtained at the endpoints with any critical points, if any. In this case, there are no critical points, so the maximum and minimum values of f(x) occur at the endpoints.
Therefore, the absolute maximum value of f(x) is 1, which occurs at x = 0 and x = π. The absolute minimum value of f(x) is also 1, which also occurs at x = 0 and x = π.
Case 2: Using the second derivative test for relative maxima and minima for f(x) = x^4/4 - 2x^2 + 1
To find the relative maxima and minima of a function, you need to find the critical points and then analyze the concavity using the second derivative.
1. Take the derivative of f(x): f'(x) = x^3 - 4x
2. Set f'(x) = 0 to find the critical points:
x^3 - 4x = 0
x(x^2 - 4) = 0
x(x - 2)(x + 2) = 0
From this, we get three critical points: x = 0, x = 2, and x = -2.
3. Take the second derivative of f(x): f''(x) = 3x^2 - 4
4. Substitute the critical points into f''(x) to analyze the concavity:
f''(0) = 3(0)^2 - 4 = -4 (negative, so it is concave down)
f''(2) = 3(2)^2 - 4 = 8 (positive, so it is concave up)
f''(-2) = 3(-2)^2 - 4 = 8 (positive, so it is concave up)
5. Now, based on the concavity, we can determine the relative extrema. If the concavity changes from concave up to concave down, we have a relative maximum; if it changes from concave down to concave up, we have a relative minimum.
- At x = -2, the concavity changes from concave up to concave down, so there is a relative maximum.
- At x = 0, there is no change in concavity, so no relative extrema.
- At x = 2, the concavity changes from concave down to concave up, so there is a relative minimum.
6. Evaluate the function at the critical points to obtain the y-coordinates of the relative extrema:
f(-2) = (-2)^4/4 - 2(-2)^2 + 1 = 6 - 8 + 1 = -1 (relative maximum)
f(0) = (0)^4/4 - 2(0)^2 + 1 = 1 (no relative extrema)
f(2) = (2)^4/4 - 2(2)^2 + 1 = 4 - 8 + 1 = -3 (relative minimum)
Therefore, there is a relative maximum at (-2, -1) and a relative minimum at (2, -3).
Case 3: Finding the values of x in [0, π] where the tangent line to the graph of f(x) = sin(x)/2 + cos(x) is horizontal.
To find the values of x where the tangent line is horizontal, we need to find the x-values where the derivative of the function is equal to 0.
Here's how you can do it step by step:
1. Take the derivative of f(x): f'(x) = (cos(x) - sin(x))/2
2. Set f'(x) = 0 to find the critical points:
(cos(x) - sin(x))/2 = 0
cos(x) - sin(x) = 0
3. Rearrange the equation to solve for x:
cos(x) = sin(x)
Divide both sides by cos(x):
1 = tan(x)
4. Solve for x by taking the inverse tangent of both sides:
x = arctan(1) = π/4
Therefore, the tangent line to the graph of f(x) = sin(x)/2 + cos(x) is horizontal at x = π/4.
I hope this explanation helps you understand how to approach these problems. If you have any further questions, feel free to ask!