Calculus Examples

$f (x) = x^{3} - 20 x^{2} + 100 x$

Step 1

Find the second derivative.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1

Differentiate.

Tap for more steps...

Step 1.1.1.1

By the Sum Rule, the derivative of $x^{3} - 20 x^{2} + 100 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 20 x^{2}] + \frac{d}{d x} [100 x]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 20 x^{2}) + \frac{d}{d x} (100 x)$

Step 1.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 3 x^{2} + \frac{d}{d x} (- 20 x^{2}) + \frac{d}{d x} (100 x)$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 20 x^{2}]$ .

Tap for more steps...

Step 1.1.2.1

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x^{2}$ with respect to $x$ is $- 20 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 x^{2} - 20 \frac{d}{d x} x^{2} + \frac{d}{d x} (100 x)$

Step 1.1.2.2

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

$f' (x) = 3 x^{2} - 20 (2 x) + \frac{d}{d x} (100 x)$

Step 1.1.2.3

Multiply $2$ by $- 20$ .

$f' (x) = 3 x^{2} - 40 x + \frac{d}{d x} (100 x)$

Step 1.1.3

Evaluate $\frac{d}{d x} [100 x]$ .

Tap for more steps...

Step 1.1.3.1

Since $100$ is constant with respect to $x$ , the derivative of $100 x$ with respect to $x$ is $100 \frac{d}{d x} [x]$ .

$f' (x) = 3 x^{2} - 40 x + 100 \frac{d}{d x} (x)$

Step 1.1.3.2

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

$f' (x) = 3 x^{2} - 40 x + 100 \cdot 1$

Step 1.1.3.3

Multiply $100$ by $1$ .

$f' (x) = 3 x^{2} - 40 x + 100$

Step 1.2

Find the second derivative.

Tap for more steps...

Step 1.2.1

By the Sum Rule, the derivative of $3 x^{2} - 40 x + 100$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 40 x] + \frac{d}{d x} [100]$ .

$f'' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 40 x) + \frac{d}{d x} (100)$

Step 1.2.2

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

Tap for more steps...

Step 1.2.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 40 x) + \frac{d}{d x} (100)$

Step 1.2.2.2

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

$f'' (x) = 3 (2 x) + \frac{d}{d x} (- 40 x) + \frac{d}{d x} (100)$

Step 1.2.2.3

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (- 40 x) + \frac{d}{d x} (100)$

Step 1.2.3

Evaluate $\frac{d}{d x} [- 40 x]$ .

Tap for more steps...

Step 1.2.3.1

Since $- 40$ is constant with respect to $x$ , the derivative of $- 40 x$ with respect to $x$ is $- 40 \frac{d}{d x} [x]$ .

$f'' (x) = 6 x - 40 \frac{d}{d x} x + \frac{d}{d x} (100)$

Step 1.2.3.2

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

$f'' (x) = 6 x - 40 \cdot 1 + \frac{d}{d x} (100)$

Step 1.2.3.3

Multiply $- 40$ by $1$ .

$f'' (x) = 6 x - 40 + \frac{d}{d x} (100)$

Step 1.2.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 1.2.4.1

Since $100$ is constant with respect to $x$ , the derivative of $100$ with respect to $x$ is $0$ .

$f'' (x) = 6 x - 40 + 0$

Step 1.2.4.2

Add $6 x - 40$ and $0$ .

$f'' (x) = 6 x - 40$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $6 x - 40$ .

$6 x - 40$

Step 2

Set the second derivative equal to $0$ then solve the equation $6 x - 40 = 0$ .

Tap for more steps...

Step 2.1

Set the second derivative equal to $0$ .

$6 x - 40 = 0$

Step 2.2

Add $40$ to both sides of the equation.

$6 x = 40$

Step 2.3

Divide each term in $6 x = 40$ by $6$ and simplify.

Tap for more steps...

Step 2.3.1

Divide each term in $6 x = 40$ by $6$ .

$\frac{6 x}{6} = \frac{40}{6}$

Step 2.3.2

Simplify the left side.

Tap for more steps...

Step 2.3.2.1

Cancel the common factor of $6$ .

Tap for more steps...

Step 2.3.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{40}{6}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{40}{6}$

Step 2.3.3

Simplify the right side.

Tap for more steps...

Step 2.3.3.1

Cancel the common factor of $40$ and $6$ .

Tap for more steps...

Step 2.3.3.1.1

Factor $2$ out of $40$ .

$x = \frac{2 (20)}{6}$

Step 2.3.3.1.2

Cancel the common factors.

Tap for more steps...

Step 2.3.3.1.2.1

Factor $2$ out of $6$ .

$x = \frac{2 \cdot 20}{2 \cdot 3}$

Step 2.3.3.1.2.2

Cancel the common factor.

$x = \frac{2 \cdot 20}{2 \cdot 3}$

Step 2.3.3.1.2.3

Rewrite the expression.

$x = \frac{20}{3}$

Step 3

Find the points where the second derivative is $0$ .

Tap for more steps...

Step 3.1

Substitute $\frac{20}{3}$ in $f (x) = x^{3} - 20 x^{2} + 100 x$ to find the value of $y$ .

Tap for more steps...

Step 3.1.1

Replace the variable $x$ with $\frac{20}{3}$ in the expression.

$f (\frac{20}{3}) = {(\frac{20}{3})}^{3} - 20 {(\frac{20}{3})}^{2} + 100 (\frac{20}{3})$

Step 3.1.2

Simplify the result.

Tap for more steps...

Step 3.1.2.1

Simplify each term.

Tap for more steps...

Step 3.1.2.1.1

Apply the product rule to $\frac{20}{3}$ .

$f (\frac{20}{3}) = \frac{20^{3}}{3^{3}} - 20 {(\frac{20}{3})}^{2} + 100 (\frac{20}{3})$

Step 3.1.2.1.2

Raise $20$ to the power of $3$ .

$f (\frac{20}{3}) = \frac{8000}{3^{3}} - 20 {(\frac{20}{3})}^{2} + 100 (\frac{20}{3})$

Step 3.1.2.1.3

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

$f (\frac{20}{3}) = \frac{8000}{27} - 20 {(\frac{20}{3})}^{2} + 100 (\frac{20}{3})$

Step 3.1.2.1.4

Apply the product rule to $\frac{20}{3}$ .

$f (\frac{20}{3}) = \frac{8000}{27} - 20 \frac{20^{2}}{3^{2}} + 100 (\frac{20}{3})$

Step 3.1.2.1.5

Raise $20$ to the power of $2$ .

$f (\frac{20}{3}) = \frac{8000}{27} - 20 \frac{400}{3^{2}} + 100 (\frac{20}{3})$

Step 3.1.2.1.6

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

$f (\frac{20}{3}) = \frac{8000}{27} - 20 (\frac{400}{9}) + 100 (\frac{20}{3})$

Step 3.1.2.1.7

Multiply $- 20 (\frac{400}{9})$ .

Tap for more steps...

Step 3.1.2.1.7.1

Combine $- 20$ and $\frac{400}{9}$ .

$f (\frac{20}{3}) = \frac{8000}{27} + \frac{- 20 \cdot 400}{9} + 100 (\frac{20}{3})$

Step 3.1.2.1.7.2

Multiply $- 20$ by $400$ .

$f (\frac{20}{3}) = \frac{8000}{27} + \frac{- 8000}{9} + 100 (\frac{20}{3})$

Step 3.1.2.1.8

Move the negative in front of the fraction.

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000}{9} + 100 (\frac{20}{3})$

Step 3.1.2.1.9

Multiply $100 (\frac{20}{3})$ .

Tap for more steps...

Step 3.1.2.1.9.1

Combine $100$ and $\frac{20}{3}$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000}{9} + \frac{100 \cdot 20}{3}$

Step 3.1.2.1.9.2

Multiply $100$ by $20$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000}{9} + \frac{2000}{3}$

Step 3.1.2.2

Find the common denominator.

Tap for more steps...

Step 3.1.2.2.1

Multiply $\frac{8000}{9}$ by $\frac{3}{3}$ .

$f (\frac{20}{3}) = \frac{8000}{27} - (\frac{8000}{9} \cdot \frac{3}{3}) + \frac{2000}{3}$

Step 3.1.2.2.2

Multiply $\frac{8000}{9}$ by $\frac{3}{3}$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000 \cdot 3}{9 \cdot 3} + \frac{2000}{3}$

Step 3.1.2.2.3

Multiply $\frac{2000}{3}$ by $\frac{9}{9}$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000 \cdot 3}{9 \cdot 3} + \frac{2000}{3} \cdot \frac{9}{9}$

Step 3.1.2.2.4

Multiply $\frac{2000}{3}$ by $\frac{9}{9}$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000 \cdot 3}{9 \cdot 3} + \frac{2000 \cdot 9}{3 \cdot 9}$

Step 3.1.2.2.5

Reorder the factors of $9 \cdot 3$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000 \cdot 3}{3 \cdot 9} + \frac{2000 \cdot 9}{3 \cdot 9}$

Step 3.1.2.2.6

Multiply $3$ by $9$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000 \cdot 3}{27} + \frac{2000 \cdot 9}{3 \cdot 9}$

Step 3.1.2.2.7

Multiply $3$ by $9$ .

$f (\frac{20}{3}) = \frac{8000}{27} - \frac{8000 \cdot 3}{27} + \frac{2000 \cdot 9}{27}$

Step 3.1.2.3

Combine the numerators over the common denominator.

$f (\frac{20}{3}) = \frac{8000 - 8000 \cdot 3 + 2000 \cdot 9}{27}$

Step 3.1.2.4

Simplify each term.

Tap for more steps...

Step 3.1.2.4.1

Multiply $- 8000$ by $3$ .

$f (\frac{20}{3}) = \frac{8000 - 24000 + 2000 \cdot 9}{27}$

Step 3.1.2.4.2

Multiply $2000$ by $9$ .

$f (\frac{20}{3}) = \frac{8000 - 24000 + 18000}{27}$

Step 3.1.2.5

Simplify by adding and subtracting.

Tap for more steps...

Step 3.1.2.5.1

Subtract $24000$ from $8000$ .

$f (\frac{20}{3}) = \frac{- 16000 + 18000}{27}$

Step 3.1.2.5.2

Add $- 16000$ and $18000$ .

$f (\frac{20}{3}) = \frac{2000}{27}$

Step 3.1.2.6

The final answer is $\frac{2000}{27}$ .

$\frac{2000}{27}$

Step 3.2

The point found by substituting $\frac{20}{3}$ in $f (x) = x^{3} - 20 x^{2} + 100 x$ is $(\frac{20}{3}, \frac{2000}{27})$ . This point can be an inflection point.

$(\frac{20}{3}, \frac{2000}{27})$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, \frac{20}{3}) \cup (\frac{20}{3}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{20}{3})$ into the second derivative to determine if it is increasing or decreasing.

Tap for more steps...

Step 5.1

Replace the variable $x$ with $6.5\overline{6}$ in the expression.

$f'' (6.5\overline{6}) = 6 (6.5\overline{6}) - 40$

Step 5.2

Simplify the result.

Tap for more steps...

Step 5.2.1

Multiply $6$ by $6.5\overline{6}$ .

$f'' (6.5\overline{6}) = 39.4 - 40$

Step 5.2.2

Subtract $40$ from $39.4$ .

$f'' (6.5\overline{6}) = - 0.5\overline{9}$

Step 5.2.3

The final answer is $- 0.5\overline{9}$ .

$- 0.5\overline{9}$

Step 5.3

At $6.5\overline{6}$ , the second derivative is $- 0.5\overline{9}$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{20}{3})$

Decreasing on $(- \infty, \frac{20}{3})$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(\frac{20}{3}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Tap for more steps...

Step 6.1

Replace the variable $x$ with $6.7\overline{6}$ in the expression.

$f'' (6.7\overline{6}) = 6 (6.7\overline{6}) - 40$

Step 6.2

Simplify the result.

Tap for more steps...

Step 6.2.1

Multiply $6$ by $6.7\overline{6}$ .

$f'' (6.7\overline{6}) = 40.6 - 40$

Step 6.2.2

Subtract $40$ from $40.6$ .

$f'' (6.7\overline{6}) = 0.6$

Step 6.2.3

The final answer is $0.6$ .

$0.6$

Step 6.3

At $6.7\overline{6}$ , the second derivative is $0.6$ . Since this is positive, the second derivative is increasing on the interval $(\frac{20}{3}, \infty)$ .

Increasing on $(\frac{20}{3}, \infty)$ since $f'' (x) > 0$

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(\frac{20}{3}, \frac{2000}{27})$ .

$(\frac{20}{3}, \frac{2000}{27})$

Step 8