Calculus Examples

$f (x) = \sqrt{4 - x}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

Tap for more steps...

Step 1.1

Find the second derivative.

Tap for more steps...

Step 1.1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 - x}$ as ${(4 - x)}^{\frac{1}{2}}$ .

$f' (x) = \frac{d}{d x} ({(4 - x)}^{\frac{1}{2}})$

Step 1.1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 4 - x$ .

Tap for more steps...

Step 1.1.1.2.1

To apply the Chain Rule, set $u$ as $4 - x$ .

$f' (x) = \frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (4 - x)$

Step 1.1.1.2.2

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

$f' (x) = \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (4 - x)$

Step 1.1.1.2.3

Replace all occurrences of $u$ with $4 - x$ .

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{\frac{1}{2} - 1} \frac{d}{d x} (4 - x)$

Step 1.1.1.3

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

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (4 - x)$

Step 1.1.1.4

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

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (4 - x)$

Step 1.1.1.5

Combine the numerators over the common denominator.

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (4 - x)$

Step 1.1.1.6

Simplify the numerator.

Tap for more steps...

Step 1.1.1.6.1

Multiply $- 1$ by $2$ .

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{\frac{1 - 2}{2}} \frac{d}{d x} (4 - x)$

Step 1.1.1.6.2

Subtract $2$ from $1$ .

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{\frac{- 1}{2}} \frac{d}{d x} (4 - x)$

Step 1.1.1.7

Combine fractions.

Tap for more steps...

Step 1.1.1.7.1

Move the negative in front of the fraction.

$f' (x) = \frac{1}{2} \cdot {(4 - x)}^{- \frac{1}{2}} \frac{d}{d x} (4 - x)$

Step 1.1.1.7.2

Combine $\frac{1}{2}$ and ${(4 - x)}^{- \frac{1}{2}}$ .

$f' (x) = \frac{{(4 - x)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (4 - x)$

Step 1.1.1.7.3

Move ${(4 - x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (4 - x)$

Step 1.1.1.8

By the Sum Rule, the derivative of $4 - x$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x]$ .

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (4) + \frac{d}{d x} (- x))$

Step 1.1.1.9

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot (0 + \frac{d}{d x} (- x))$

Step 1.1.1.10

Add $0$ and $\frac{d}{d x} [- x]$ .

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (- x)$

Step 1.1.1.11

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

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot (- \frac{d}{d x} x)$

Step 1.1.1.12

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

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot (- 1 \cdot 1)$

Step 1.1.1.13

Combine fractions.

Tap for more steps...

Step 1.1.1.13.1

Multiply $- 1$ by $1$ .

$f' (x) = \frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \cdot - 1$

Step 1.1.1.13.2

Combine $\frac{1}{2 {(4 - x)}^{\frac{1}{2}}}$ and $- 1$ .

$f' (x) = \frac{- 1}{2 {(4 - x)}^{\frac{1}{2}}}$

Step 1.1.1.13.3

Move the negative in front of the fraction.

$f' (x) = - \frac{1}{2 {(4 - x)}^{\frac{1}{2}}}$

Step 1.1.2

Find the second derivative.

Tap for more steps...

Step 1.1.2.1

Differentiate using the Constant Multiple Rule.

Tap for more steps...

Step 1.1.2.1.1

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{1}{2 {(4 - x)}^{\frac{1}{2}}}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [\frac{1}{{(4 - x)}^{\frac{1}{2}}}]$ .

$f'' (x) = - \frac{1}{2} \cdot \frac{d}{d x} \frac{1}{{(4 - x)}^{\frac{1}{2}}}$

Step 1.1.2.1.2

Apply basic rules of exponents.

Tap for more steps...

Step 1.1.2.1.2.1

Rewrite $\frac{1}{{(4 - x)}^{\frac{1}{2}}}$ as ${({(4 - x)}^{\frac{1}{2}})}^{- 1}$ .

$f'' (x) = - \frac{1}{2} \cdot \frac{d}{d x} {({(4 - x)}^{\frac{1}{2}})}^{- 1}$

Step 1.1.2.1.2.2

Multiply the exponents in ${({(4 - x)}^{\frac{1}{2}})}^{- 1}$ .

Tap for more steps...

Step 1.1.2.1.2.2.1

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

$f'' (x) = - \frac{1}{2} \cdot \frac{d}{d x} {(4 - x)}^{\frac{1}{2} \cdot - 1}$

Step 1.1.2.1.2.2.2

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

$f'' (x) = - \frac{1}{2} \cdot \frac{d}{d x} {(4 - x)}^{\frac{- 1}{2}}$

Step 1.1.2.1.2.2.3

Move the negative in front of the fraction.

$f'' (x) = - \frac{1}{2} \cdot \frac{d}{d x} {(4 - x)}^{- \frac{1}{2}}$

Step 1.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{1}{2}}$ and $g (x) = 4 - x$ .

Tap for more steps...

Step 1.1.2.2.1

To apply the Chain Rule, set $u$ as $4 - x$ .

$f'' (x) = - \frac{1}{2} \cdot (\frac{d}{d u} (u^{- \frac{1}{2}}) \frac{d}{d x} (4 - x))$

Step 1.1.2.2.2

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

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} u^{- \frac{1}{2} - 1} \frac{d}{d x} (4 - x))$

Step 1.1.2.2.3

Replace all occurrences of $u$ with $4 - x$ .

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{- \frac{1}{2} - 1} \frac{d}{d x} (4 - x))$

Step 1.1.2.3

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

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{- \frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (4 - x))$

Step 1.1.2.4

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

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{- \frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (4 - x))$

Step 1.1.2.5

Combine the numerators over the common denominator.

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{\frac{- 1 - 1 \cdot 2}{2}} \frac{d}{d x} (4 - x))$

Step 1.1.2.6

Simplify the numerator.

Tap for more steps...

Step 1.1.2.6.1

Multiply $- 1$ by $2$ .

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{\frac{- 1 - 2}{2}} \frac{d}{d x} (4 - x))$

Step 1.1.2.6.2

Subtract $2$ from $- 1$ .

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{\frac{- 3}{2}} \frac{d}{d x} (4 - x))$

Step 1.1.2.7

Combine fractions.

Tap for more steps...

Step 1.1.2.7.1

Move the negative in front of the fraction.

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(4 - x)}^{- \frac{3}{2}} \frac{d}{d x} (4 - x))$

Step 1.1.2.7.2

Combine ${(4 - x)}^{- \frac{3}{2}}$ and $\frac{1}{2}$ .

$f'' (x) = - \frac{1}{2} \cdot (- \frac{{(4 - x)}^{- \frac{3}{2}}}{2} \cdot \frac{d}{d x} (4 - x))$

Step 1.1.2.7.3

Simplify the expression.

Tap for more steps...

Step 1.1.2.7.3.1

Move ${(4 - x)}^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = - \frac{1}{2} \cdot (- \frac{1}{2 {(4 - x)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (4 - x))$

Step 1.1.2.7.3.2

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (\frac{1}{2}) \cdot (\frac{1}{2 {(4 - x)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (4 - x))$

Step 1.1.2.7.3.3

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

$f'' (x) = \frac{1}{2} \cdot (\frac{1}{2 {(4 - x)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (4 - x))$

Step 1.1.2.7.4

Multiply $\frac{1}{2 {(4 - x)}^{\frac{3}{2}}}$ by $\frac{1}{2}$ .

$f'' (x) = \frac{1}{2 {(4 - x)}^{\frac{3}{2}} \cdot 2} \cdot \frac{d}{d x} (4 - x)$

Step 1.1.2.7.5

Multiply $2$ by $2$ .

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (4 - x)$

Step 1.1.2.8

By the Sum Rule, the derivative of $4 - x$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x]$ .

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot (\frac{d}{d x} (4) + \frac{d}{d x} (- x))$

Step 1.1.2.9

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot (0 + \frac{d}{d x} (- x))$

Step 1.1.2.10

Add $0$ and $\frac{d}{d x} [- x]$ .

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (- x)$

Step 1.1.2.11

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

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot (- \frac{d}{d x} x)$

Step 1.1.2.12

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

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot (- 1 \cdot 1)$

Step 1.1.2.13

Combine fractions.

Tap for more steps...

Step 1.1.2.13.1

Multiply $- 1$ by $1$ .

$f'' (x) = \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} \cdot - 1$

Step 1.1.2.13.2

Combine $\frac{1}{4 {(4 - x)}^{\frac{3}{2}}}$ and $- 1$ .

$f'' (x) = \frac{- 1}{4 {(4 - x)}^{\frac{3}{2}}}$

Step 1.1.2.13.3

Move the negative in front of the fraction.

$f'' (x) = - \frac{1}{4 {(4 - x)}^{\frac{3}{2}}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{1}{4 {(4 - x)}^{\frac{3}{2}}}$ .

$- \frac{1}{4 {(4 - x)}^{\frac{3}{2}}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} = 0$ .

Tap for more steps...

Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{1}{4 {(4 - x)}^{\frac{3}{2}}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$1 = 0$

Step 1.2.3

Since $1 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = \sqrt{4 - x}$ .

Tap for more steps...

Step 2.1

Set the radicand in $\sqrt{4 - x}$ greater than or equal to $0$ to find where the expression is defined.

$4 - x \geq 0$

Step 2.2

Solve for $x$ .

Tap for more steps...

Step 2.2.1

Subtract $4$ from both sides of the inequality.

$- x \geq - 4$

Step 2.2.2

Divide each term in $- x \geq - 4$ by $- 1$ and simplify.

Tap for more steps...

Step 2.2.2.1

Divide each term in $- x \geq - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 4}{- 1}$

Step 2.2.2.2

Simplify the left side.

Tap for more steps...

Step 2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 4}{- 1}$

Step 2.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 4}{- 1}$

Step 2.2.2.3

Simplify the right side.

Tap for more steps...

Step 2.2.2.3.1

Divide $- 4$ by $- 1$ .

$x \leq 4$

Step 2.3

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

Interval Notation:

$(- \infty, 4]$

Set-Builder Notation:

${x | x \leq 4}$

Interval Notation:

$(- \infty, 4]$

Set-Builder Notation:

${x | x \leq 4}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, 4]$

Step 4

Substitute any number from the interval $(- \infty, 4]$ into the second derivative and evaluate to determine the concavity.

Tap for more steps...

Step 4.1

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

$f'' (0) = - \frac{1}{4 {(4 - (0))}^{\frac{3}{2}}}$

Step 4.2

Simplify the result.

Tap for more steps...

Step 4.2.1

Simplify the denominator.

Tap for more steps...

Step 4.2.1.1

Subtract $0$ from $4$ .

$f'' (0) = - \frac{1}{4 \cdot 4^{\frac{3}{2}}}$

Step 4.2.1.2

Rewrite $4$ as $2^{2}$ .

$f'' (0) = - \frac{1}{4 \cdot {(2^{2})}^{\frac{3}{2}}}$

Step 4.2.1.3

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

$f'' (0) = - \frac{1}{4 \cdot 2^{2 (\frac{3}{2})}}$

Step 4.2.1.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.2.1.4.1

Cancel the common factor.

$f'' (0) = - \frac{1}{4 \cdot 2^{2 (\frac{3}{2})}}$

Step 4.2.1.4.2

Rewrite the expression.

$f'' (0) = - \frac{1}{4 \cdot 2^{3}}$

Step 4.2.1.5

Raise $2$ to the power of $3$ .

$f'' (0) = - \frac{1}{4 \cdot 8}$

Step 4.2.2

Multiply $4$ by $8$ .

$f'' (0) = - \frac{1}{32}$

Step 4.2.3

The final answer is $- \frac{1}{32}$ .

$- \frac{1}{32}$

Step 4.3

The graph is concave down on the interval $(- \infty, 4]$ because $f'' (0)$ is negative.

Concave down on $(- \infty, 4]$ since $f'' (x)$ is negative

Step 5