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(1)

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Step 1:

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Critical number :

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A critical number of a function $\"\"$ is a number $\"\"$ in the domain of $\"\"$ such that either $\"\"$ or $\"\"$ does not exist.

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The function is $\"image\"$ .

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$\"\"$ is continuous and differentiable at all values of $\"\"$ because it is a polynomial.

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Solutions of $\"\"$ are the critical numbers.

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Differentiate $\"\"$ on each side with respect to $\"\"$ .

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$\"image\"$

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$\"image\"$ .

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Step 2:

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$\"image\"$ .

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Equate $\"\"$ to zero.

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$

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$\"image\"$ .

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Critical number is $\"image\"$ .

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Solution:

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Critical number is $\"image\"$ .

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(2)

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Critical number :

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A critical number of a function $\"\"$ is a number $\"\"$ in the domain of $\"\"$ such that either $\"\"$ or $\"\"$ does not exist.

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The function is $\"\"$ .

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The domain of a function is all values of $\"\"$ , those makes the function mathematically correct.

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There should not be any negative number in the square root.

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$\"\"$

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$\"\"$

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$\"\"$ .

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The domain of the function is $\"\"$ .

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Solutions of $\"\"$ are the critical numbers.

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Step 2:

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Differentiate $\"\"$ on each side with respect to $\"\"$ .

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$\"\"$

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Apply product rule in differentiation: $\"\"$ .

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$\"\"$

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$\"\"$ .

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Step 3:

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$\"\"$ .

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Equate $\"\"$ to zero.

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$ .

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$\"\"$ is not defined at $\"\"$ . \ \

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$\"\"$ is in the domain of $\"\"$ .

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The critical points are $\"\"$ and $\"\"$ .

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Solution:

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The critical points are $\"\"$ and $\"\"$ .

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(3)

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Step 1:

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The function is $\"\"$ , on the interval $\"\"$ . \ \

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Absolute values of a function exist either at the end points or at the critical points.

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Evaluate the critical points.

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The function is $\"\"$ .

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Differentiate $\"\"$ on each side with respect to $\"\"$ .

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$\"\"$

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$\"\"$ .

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Find the critical points, equate $\"\"$ to zero.

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$ .

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Step 2: \ \

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Find the maximum and minimum values.

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Substitute the critical point in the function.

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$ \ \

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$\"\"$ .

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Evaluate function at the end points.

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The function is $\"\"$ on the interval $\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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$\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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$\"\"$ .

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The maximum value of the function is at $\"\"$ .

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The minimum value of the function is at $\"\"$ .

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The absolute maximum is $\"\"$ .

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The absolute minimum is $\"\"$ .

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Solution: \ \

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The absolute maximum is $\"\"$ .

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The absolute minimum is $\"\"$ . \ \

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(4)

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Step 1:

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The function is $\"\"$ , on the interval $\"\"$ . \ \

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Absolute values of a function exist either at the end points or at the critical points.

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Evaluate the critical points.

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The function is $\"\"$ .

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Differentiate $\"\"$ on each side with respect to $\"\"$ .

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$\"\"$

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$\"\"$ .

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Find the critical points, by equate $\"\"$ to zero.

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$\"\"$

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$\"\"$

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Solution of the equation in the interval $\"\"$ is $\"\"$ .

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the critical point is $\"\"$ .

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Step 2: \ \

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Find the maximum and minimum values.

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Substitute the critical point in the function.

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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$\"\"$ .

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Evaluate function at the end points.

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The function is $\"\"$ on the interval $\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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$\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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$\"\"$ .

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The maximum value of the function is at $\"\"$ .

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The absolute maximum is $\"\"$ .

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The minimum value of the function is at $\"\"$ . \ \

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The absolute minimum is $\"\"$ .

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Solution: \ \

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The absolute maximum is $\"\"$ .

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The absolute minimum is $\"\"$ .

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(5) \ \

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The function is $\"\"$ and the point is $\"\"$ .

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Apply derivative on each side with respect to $\"\"$ . \ \

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$\"\"$

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$\"\"$ .

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Find the slope of a tangent at the point $\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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$\"\"$ .

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Step 2:

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Find the tangent line equation.

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Point - slope form of line equation is $\"image\"$ .

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Substitute the values $\"\"$ and $\"\"$ in point slope form. \ \

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$\"\"$

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$\"\"$

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$\"\"$

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$\"\"$

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The tangent line equation is $\"\"$ .

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Solution:

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The tangent line equation is $\"\"$ .

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