Calculus Examples

$f (x) = \sqrt{x} - \frac{1}{9} x$ , $[0, 81]$

Step 1

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 $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

Check if $f (x) = \sqrt{x} - \frac{1}{9} x$ is continuous.

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Step 2.1

To find whether the function is continuous on $[0, 81]$ or not, find the domain of $f (x) = \sqrt{x} - \frac{1}{9} x$ .

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Step 2.1.1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2.1.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 2.2

$f (x)$ is continuous on $[0, 81]$ .

The function is continuous.

Step 3

Find the derivative.

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Step 3.1

Find the first derivative.

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Step 3.1.1

By the Sum Rule, the derivative of $\sqrt{x} - \frac{1}{9} x$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x}] + \frac{d}{d x} [- \frac{1}{9} x]$ .

$\frac{d}{d x} [\sqrt{x}] + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2

Evaluate $\frac{d}{d x} [\sqrt{x}]$ .

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Step 3.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$\frac{d}{d x} [x^{\frac{1}{2}}] + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = \frac{1}{2}$ .

$\frac{1}{2} x^{\frac{1}{2} - 1} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.4

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.5

Combine the numerators over the common denominator.

$\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.6

Simplify the numerator.

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Step 3.1.2.6.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} x^{\frac{1 - 2}{2}} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.6.2

Subtract $2$ from $1$ .

$\frac{1}{2} x^{\frac{- 1}{2}} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.2.7

Move the negative in front of the fraction.

$\frac{1}{2} x^{- \frac{1}{2}} + \frac{d}{d x} [- \frac{1}{9} x]$

Step 3.1.3

Evaluate $\frac{d}{d x} [- \frac{1}{9} x]$ .

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Step 3.1.3.1

Since $- \frac{1}{9}$ is constant with respect to $x$ , the derivative of $- \frac{1}{9} x$ with respect to $x$ is $- \frac{1}{9} \frac{d}{d x} [x]$ .

$\frac{1}{2} x^{- \frac{1}{2}} - \frac{1}{9} \frac{d}{d x} [x]$

Step 3.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{2} x^{- \frac{1}{2}} - \frac{1}{9} \cdot 1$

Step 3.1.3.3

Multiply $- 1$ by $1$ .

$\frac{1}{2} x^{- \frac{1}{2}} - \frac{1}{9}$

Step 3.1.4

Simplify.

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Step 3.1.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2} \cdot \frac{1}{x^{\frac{1}{2}}} - \frac{1}{9}$

Step 3.1.4.2

Multiply $\frac{1}{2}$ by $\frac{1}{x^{\frac{1}{2}}}$ .

$f' (x) = \frac{1}{2 x^{\frac{1}{2}}} - \frac{1}{9}$

Step 3.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{1}{2 x^{\frac{1}{2}}} - \frac{1}{9}$ .

$\frac{1}{2 x^{\frac{1}{2}}} - \frac{1}{9}$

Step 4

Find if the derivative is continuous on $(0, 81)$ .

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Step 4.1

To find whether the function is continuous on $(0, 81)$ or not, find the domain of $f' (x) = \frac{1}{2 x^{\frac{1}{2}}} - \frac{1}{9}$ .

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Step 4.1.1

Convert expressions with fractional exponents to radicals.

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Step 4.1.1.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{1}{2 \sqrt{x^{1}}} - \frac{1}{9}$

Step 4.1.1.2

Anything raised to $1$ is the base itself.

$\frac{1}{2 \sqrt{x}} - \frac{1}{9}$

Step 4.1.2

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 4.1.3

Set the denominator in $\frac{1}{2 \sqrt{x}}$ equal to $0$ to find where the expression is undefined.

$2 \sqrt{x} = 0$

Step 4.1.4

Solve for $x$ .

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Step 4.1.4.1

To remove the radical on the left side of the equation, square both sides of the equation.

${(2 \sqrt{x})}^{2} = 0^{2}$

Step 4.1.4.2

Simplify each side of the equation.

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Step 4.1.4.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(2 x^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.1.4.2.2

Simplify the left side.

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Step 4.1.4.2.2.1

Simplify ${(2 x^{\frac{1}{2}})}^{2}$ .

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Step 4.1.4.2.2.1.1

Apply the product rule to $2 x^{\frac{1}{2}}$ .

$2^{2} {(x^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.1.4.2.2.1.2

Raise $2$ to the power of $2$ .

$4 {(x^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.1.4.2.2.1.3

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 4.1.4.2.2.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 x^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.1.4.2.2.1.3.2

Cancel the common factor of $2$ .

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Step 4.1.4.2.2.1.3.2.1

Cancel the common factor.

$4 x^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.1.4.2.2.1.3.2.2

Rewrite the expression.

$4 x^{1} = 0^{2}$

Step 4.1.4.2.2.1.4

Simplify.

$4 x = 0^{2}$

Step 4.1.4.2.3

Simplify the right side.

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Step 4.1.4.2.3.1

Raising $0$ to any positive power yields $0$ .

$4 x = 0$

Step 4.1.4.3

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 4.1.4.3.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 4.1.4.3.2

Simplify the left side.

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Step 4.1.4.3.2.1

Cancel the common factor of $4$ .

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Step 4.1.4.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 4.1.4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 4.1.4.3.3

Simplify the right side.

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Step 4.1.4.3.3.1

Divide $0$ by $4$ .

$x = 0$

Step 4.1.5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 4.2

$f' (x)$ is continuous on $(0, 81)$ .

The function is continuous.

Step 5

The function is differentiable on $(0, 81)$ because the derivative is continuous on $(0, 81)$ .

The function is differentiable.

Step 6

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[0, 81]$ and differentiable on $(0, 81)$ .

$f (x)$ is continuous on $[0, 81]$ and differentiable on $(0, 81)$ .

Step 7

Evaluate $f (a)$ from the interval $[0, 81]$ .

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Step 7.1

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

$f (0) = \sqrt{0} - \frac{1}{9} \cdot 0$

Step 7.2

Simplify the result.

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Step 7.2.1

Remove parentheses.

$f (0) = \sqrt{0} - \frac{1}{9} \cdot 0$

Step 7.2.2

Simplify each term.

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Step 7.2.2.1

Rewrite $0$ as $0^{2}$ .

$f (0) = \sqrt{0^{2}} - \frac{1}{9} \cdot 0$

Step 7.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = 0 - \frac{1}{9} \cdot 0$

Step 7.2.2.3

Multiply $- \frac{1}{9} \cdot 0$ .

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Step 7.2.2.3.1

Multiply $0$ by $- 1$ .

$f (0) = 0 + 0 (\frac{1}{9})$

Step 7.2.2.3.2

Multiply $0$ by $\frac{1}{9}$ .

$f (0) = 0 + 0$

Step 7.2.3

Add $0$ and $0$ .

$f (0) = 0$

Step 7.2.4

The final answer is $0$ .

$0$

Step 8

Solve $\frac{1}{2 x^{\frac{1}{2}}} - \frac{1}{9} = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $\frac{1}{2 x^{\frac{1}{2}}} - \frac{1}{9} = \frac{- (f (81) + f (0))}{- (81 + 0)}$ .

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Step 8.1

Move all terms not containing $x$ to the right side of the equation.

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Step 8.1.1

Add $\frac{1}{9}$ to both sides of the equation.

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{(0) - (0)}{(81) - (0)} + \frac{1}{9}$

Step 8.1.2

Multiply $- 1$ by $0$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0 + 0}{(81) - (0)} + \frac{1}{9}$

Step 8.1.3

Add $0$ and $0$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0}{(81) - (0)} + \frac{1}{9}$

Step 8.1.4

Simplify each term.

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Step 8.1.4.1

Cancel the common factor of $0$ and $(81) - (0)$ .

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Step 8.1.4.1.1

Rewrite $81$ as $- 1 (- 81)$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0}{- 1 (- 81) - (0)} + \frac{1}{9}$

Step 8.1.4.1.2

Factor $- 1$ out of $- 1 (- 81) - (0)$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0}{- 1 (- 81 + 0)} + \frac{1}{9}$

Step 8.1.4.1.3

Factor $81$ out of $0$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{81 (0)}{- 1 (- 81 + 0)} + \frac{1}{9}$

Step 8.1.4.1.4

Cancel the common factors.

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Step 8.1.4.1.4.1

Factor $81$ out of $- 1 (- 81 + 0)$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{81 (0)}{81 (- 1 (- 1 + 0))} + \frac{1}{9}$

Step 8.1.4.1.4.2

Cancel the common factor.

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{81 \cdot 0}{81 (- 1 (- 1 + 0))} + \frac{1}{9}$

Step 8.1.4.1.4.3

Rewrite the expression.

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0}{- 1 (- 1 + 0)} + \frac{1}{9}$

Step 8.1.4.2

Add $- 1$ and $0$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0}{- 1 \cdot - 1} + \frac{1}{9}$

Step 8.1.4.3

Multiply $- 1$ by $- 1$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{0}{1} + \frac{1}{9}$

Step 8.1.4.4

Divide $0$ by $1$ .

$\frac{1}{2 x^{\frac{1}{2}}} = 0 + \frac{1}{9}$

Step 8.1.5

Add $0$ and $\frac{1}{9}$ .

$\frac{1}{2 x^{\frac{1}{2}}} = \frac{1}{9}$

Step 8.2

Find the LCD of the terms in the equation.

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Step 8.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 x^{\frac{1}{2}}, 9$

Step 8.2.2

Since $2 x^{\frac{1}{2}}, 9$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $2, 9$ then find LCM for the variable part $x^{\frac{1}{2}}$ .

Step 8.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 8.2.4

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 8.2.5

$9$ has factors of $3$ and $3$ .

$3 \cdot 3$

Step 8.2.6

The LCM of $2, 9$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 3 \cdot 3$

Step 8.2.7

Multiply $2 \cdot 3 \cdot 3$ .

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Step 8.2.7.1

Multiply $2$ by $3$ .

$6 \cdot 3$

Step 8.2.7.2

Multiply $6$ by $3$ .

$18$

Step 8.2.8

The LCM of $x^{\frac{1}{2}}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x^{\frac{1}{2}}$

Step 8.2.9

The LCM for $2 x^{\frac{1}{2}}, 9$ is the numeric part $18$ multiplied by the variable part.

$18 x^{\frac{1}{2}}$

Step 8.3

Multiply each term in $\frac{1}{2 x^{\frac{1}{2}}} = \frac{1}{9}$ by $18 x^{\frac{1}{2}}$ to eliminate the fractions.

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Step 8.3.1

Multiply each term in $\frac{1}{2 x^{\frac{1}{2}}} = \frac{1}{9}$ by $18 x^{\frac{1}{2}}$ .

$\frac{1}{2 x^{\frac{1}{2}}} (18 x^{\frac{1}{2}}) = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2

Simplify the left side.

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Step 8.3.2.1

Rewrite using the commutative property of multiplication.

$18 \frac{1}{2 x^{\frac{1}{2}}} x^{\frac{1}{2}} = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2.2

Cancel the common factor of $2$ .

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Step 8.3.2.2.1

Factor $2$ out of $18$ .

$2 \cdot 9 \frac{1}{2 x^{\frac{1}{2}}} x^{\frac{1}{2}} = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2.2.2

Cancel the common factor.

$2 \cdot 9 \frac{1}{2 x^{\frac{1}{2}}} x^{\frac{1}{2}} = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2.2.3

Rewrite the expression.

$9 \frac{1}{x^{\frac{1}{2}}} x^{\frac{1}{2}} = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2.3

Combine $9$ and $\frac{1}{x^{\frac{1}{2}}}$ .

$\frac{9}{x^{\frac{1}{2}}} x^{\frac{1}{2}} = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2.4

Cancel the common factor of $x^{\frac{1}{2}}$ .

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Step 8.3.2.4.1

Cancel the common factor.

$\frac{9}{x^{\frac{1}{2}}} x^{\frac{1}{2}} = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.2.4.2

Rewrite the expression.

$9 = \frac{1}{9} (18 x^{\frac{1}{2}})$

Step 8.3.3

Simplify the right side.

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Step 8.3.3.1

Cancel the common factor of $9$ .

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Step 8.3.3.1.1

Factor $9$ out of $18 x^{\frac{1}{2}}$ .

$9 = \frac{1}{9} (9 (2 x^{\frac{1}{2}}))$

Step 8.3.3.1.2

Cancel the common factor.

$9 = \frac{1}{9} (9 (2 x^{\frac{1}{2}}))$

Step 8.3.3.1.3

Rewrite the expression.

$9 = 2 x^{\frac{1}{2}}$

Step 8.4

Solve the equation.

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Step 8.4.1

Rewrite the equation as $2 x^{\frac{1}{2}} = 9$ .

$2 x^{\frac{1}{2}} = 9$

Step 8.4.2

Divide each term in $2 x^{\frac{1}{2}} = 9$ by $2$ and simplify.

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Step 8.4.2.1

Divide each term in $2 x^{\frac{1}{2}} = 9$ by $2$ .

$\frac{2 x^{\frac{1}{2}}}{2} = \frac{9}{2}$

Step 8.4.2.2

Simplify the left side.

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Step 8.4.2.2.1

Cancel the common factor.

$\frac{2 x^{\frac{1}{2}}}{2} = \frac{9}{2}$

Step 8.4.2.2.2

Divide $x^{\frac{1}{2}}$ by $1$ .

$x^{\frac{1}{2}} = \frac{9}{2}$

Step 8.4.3

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

${(x^{\frac{1}{2}})}^{2} = {(\frac{9}{2})}^{2}$

Step 8.4.4

Simplify the exponent.

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Step 8.4.4.1

Simplify the left side.

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Step 8.4.4.1.1

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

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Step 8.4.4.1.1.1

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 8.4.4.1.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} = {(\frac{9}{2})}^{2}$

Step 8.4.4.1.1.1.2

Cancel the common factor of $2$ .

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Step 8.4.4.1.1.1.2.1

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} = {(\frac{9}{2})}^{2}$

Step 8.4.4.1.1.1.2.2

Rewrite the expression.

$x^{1} = {(\frac{9}{2})}^{2}$

Step 8.4.4.1.1.2

Simplify.

$x = {(\frac{9}{2})}^{2}$

Step 8.4.4.2

Simplify the right side.

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Step 8.4.4.2.1

Simplify ${(\frac{9}{2})}^{2}$ .

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Step 8.4.4.2.1.1

Apply the product rule to $\frac{9}{2}$ .

$x = \frac{9^{2}}{2^{2}}$

Step 8.4.4.2.1.2

Raise $9$ to the power of $2$ .

$x = \frac{81}{2^{2}}$

Step 8.4.4.2.1.3

Raise $2$ to the power of $2$ .

$x = \frac{81}{4}$

Step 9

There is a tangent line found at $x = \frac{81}{4}$ parallel to the line that passes through the end points $a = 0$ and $b = 81$ .

There is a tangent line at $x = \frac{81}{4}$ parallel to the line that passes through the end points $a = 0$ and $b = 81$

Step 10