Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=cos(x)
Step 1
The derivative of with respect to is .
Step 2
Find the second derivative of the function.
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Step 2.1
Since is constant with respect to , the derivative of with respect to is .
Step 2.2
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
Divide each term in by and simplify.
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Step 4.1
Divide each term in by .
Step 4.2
Simplify the left side.
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Step 4.2.1
Dividing two negative values results in a positive value.
Step 4.2.2
Divide by .
Step 4.3
Simplify the right side.
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Step 4.3.1
Divide by .
Step 5
Take the inverse sine of both sides of the equation to extract from inside the sine.
Step 6
Simplify the right side.
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Step 6.1
The exact value of is .
Step 7
The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from to find the solution in the second quadrant.
Step 8
Subtract from .
Step 9
The solution to the equation .
Step 10
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 11
Evaluate the second derivative.
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Step 11.1
The exact value of is .
Step 11.2
Multiply by .
Step 12
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 13
Find the y-value when .
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Step 13.1
Replace the variable with in the expression.
Step 13.2
Simplify the result.
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Step 13.2.1
The exact value of is .
Step 13.2.2
The final answer is .
Step 14
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 15
Evaluate the second derivative.
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Step 15.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.
Step 15.2
The exact value of is .
Step 15.3
Multiply .
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Step 15.3.1
Multiply by .
Step 15.3.2
Multiply by .
Step 16
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 17
Find the y-value when .
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Step 17.1
Replace the variable with in the expression.
Step 17.2
Simplify the result.
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Step 17.2.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.
Step 17.2.2
The exact value of is .
Step 17.2.3
Multiply by .
Step 17.2.4
The final answer is .
Step 18
These are the local extrema for .
is a local maxima
is a local minima
Step 19