Calculus Examples

$f (x) = 8 \sin (x)$ , $[0, 2 π]$

If $f$ is continuous on the interval $[a, b]$ and differentiable on $(a, b)$ , then at least one real number $c$ exists in the interval $(a, b)$ such that $f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at $x = c$ and the slope of the line through the points $(a, f (a))$ and $(b, f (b))$ .

If $f (x)$ is continuous on $[a, b]$

and if $f (x)$ differentiable on $(a, b)$ ,

then there exists at least one point, $c$ in $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Check if $f (x) = 8 \sin (x)$ is continuous.

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

$f (x)$ is continuous on $[0, 2 π]$ .

The function is continuous.

Find the derivative.

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Find the first derivative.

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Since $8$ is constant with respect to $x$ , the derivative of $8 \sin (x)$ with respect to $x$ is $8 \frac{d}{d x} [\sin (x)]$ .

$f' (x) = 8 \frac{d}{d x} (\sin (x))$

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f' (x) = 8 \cos (x)$

The first derivative of $f (x)$ with respect to $x$ is $8 \cos (x)$ .

$8 \cos (x)$

Find if the derivative is continuous on $(0, 2 π)$ .

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

$f' (x)$ is continuous on $(0, 2 π)$ .

The function is continuous.

The function is differentiable on $(0, 2 π)$ because the derivative is continuous on $(0, 2 π)$ .

The function is differentiable.

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[0, 2 π]$ and differentiable on $(0, 2 π)$ .

$f (x)$ is continuous on $[0, 2 π]$ and differentiable on $(0, 2 π)$ .

Evaluate $f (a)$ from the interval $[0, 2 π]$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = 8 \sin (0)$

Simplify the result.

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The exact value of $\sin (0)$ is $0$ .

$f (0) = 8 \cdot 0$

Multiply $8$ by $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

Solve $8 \cos (x) = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $8 \cos (x) = \frac{- (f (2 π) + f (0))}{- (2 π + 0)}$ .

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Simplify.

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Multiply $- 1$ by $0$ .

$8 \cos (x) = \frac{0 + 0}{2 π - (0)}$

Multiply $- 1$ by $0$ .

$8 \cos (x) = \frac{0 + 0}{2 π + 0}$

Reduce the expression $\frac{0 + 0}{2 π + 0}$ by cancelling the common factors.

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Factor $2$ out of $0$ .

$8 \cos (x) = \frac{2 (0) + 0}{2 π + 0}$

Factor $2$ out of $0$ .

$8 \cos (x) = \frac{2 \cdot 0 + 2 \cdot 0}{2 π + 0}$

Factor $2$ out of $2 \cdot 0 + 2 \cdot 0$ .

$8 \cos (x) = \frac{2 \cdot (0 + 0)}{2 π + 0}$

Factor $2$ out of $2 π$ .

$8 \cos (x) = \frac{2 \cdot (0 + 0)}{2 (π) + 0}$

Factor $2$ out of $0$ .

$8 \cos (x) = \frac{2 \cdot (0 + 0)}{2 (π) + 2 (0)}$

Factor $2$ out of $2 (π) + 2 (0)$ .

$8 \cos (x) = \frac{2 \cdot (0 + 0)}{2 (π + 0)}$

Cancel the common factor.

$8 \cos (x) = \frac{2 \cdot (0 + 0)}{2 (π + 0)}$

Rewrite the expression.

$8 \cos (x) = \frac{0 + 0}{π + 0}$

Add $0$ and $0$ .

$8 \cos (x) = \frac{0}{π + 0}$

Add $π$ and $0$ .

$8 \cos (x) = \frac{0}{π}$

Divide $0$ by $π$ .

$8 \cos (x) = 0$

Divide each term in $8 \cos (x) = 0$ by $8$ and simplify.

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Divide each term in $8 \cos (x) = 0$ by $8$ .

$\frac{8 \cos (x)}{8} = \frac{0}{8}$

Simplify the left side.

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Cancel the common factor of $8$ .

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Cancel the common factor.

$\frac{8 \cos (x)}{8} = \frac{0}{8}$

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{0}{8}$

Simplify the right side.

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Divide $0$ by $8$ .

$\cos (x) = 0$

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$x = \arccos (0)$

Simplify the right side.

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The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

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Combine $2 π$ and $\frac{2}{2}$ .

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

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Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Find the period of $\cos (x)$ .

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The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Consolidate the answers.

$x = \frac{π}{2} + π n$ , for any integer $n$

There is a tangent line found at $x = \frac{π}{2} + π n$ parallel to the line that passes through the end points $a = 0$ and $b = 2 π$ .

There is a tangent line at $x = \frac{π}{2} + π n$ parallel to the line that passes through the end points $a = 0$ and $b = 2 π$