Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=sin(x)
Step 1
The derivative of with respect to is .
Step 2
The derivative of with respect to is .
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to and solve.
Step 4
Take the inverse cosine of both sides of the equation to extract from inside the cosine.
Step 5
Simplify the right side.
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Step 5.1
The exact value of is .
Step 6
The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from to find the solution in the fourth quadrant.
Step 7
Simplify .
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Step 7.1
To write as a fraction with a common denominator, multiply by .
Step 7.2
Combine fractions.
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Step 7.2.1
Combine and .
Step 7.2.2
Combine the numerators over the common denominator.
Step 7.3
Simplify the numerator.
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Step 7.3.1
Multiply by .
Step 7.3.2
Subtract from .
Step 8
The solution to the equation .
Step 9
Evaluate the second derivative at . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
Step 10
Evaluate the second derivative.
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Step 10.1
The exact value of is .
Step 10.2
Multiply by .
Step 11
is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
is a local maximum
Step 12
Find the y-value when .
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Step 12.1
Replace the variable with in the expression.
Step 12.2
Simplify the result.
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Step 12.2.1
The exact value of is .
Step 12.2.2
The final answer is .
Step 13
Evaluate the second derivative at . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
Step 14
Evaluate the second derivative.
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Step 14.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.
Step 14.2
The exact value of is .
Step 14.3
Multiply .
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Step 14.3.1
Multiply by .
Step 14.3.2
Multiply by .
Step 15
is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
is a local minimum
Step 16
Find the y-value when .
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Step 16.1
Replace the variable with in the expression.
Step 16.2
Simplify the result.
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Step 16.2.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.
Step 16.2.2
The exact value of is .
Step 16.2.3
Multiply by .
Step 16.2.4
The final answer is .
Step 17
These are the local extrema for .
is a local maxima
is a local minima
Step 18