Calculus Examples

$f (x) = 6 x^{6} - 13 x^{5}$

Step 1

Find the second derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $6 x^{6} - 13 x^{5}$ with respect to $x$ is $\frac{d}{d x} [6 x^{6}] + \frac{d}{d x} [- 13 x^{5}]$ .

$f' (x) = \frac{d}{d x} (6 x^{6}) + \frac{d}{d x} (- 13 x^{5})$

Evaluate $\frac{d}{d x} [6 x^{6}]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x^{6}$ with respect to $x$ is $6 \frac{d}{d x} [x^{6}]$ .

$f' (x) = 6 \frac{d}{d x} (x^{6}) + \frac{d}{d x} (- 13 x^{5})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 6$ .

$f' (x) = 6 (6 x^{5}) + \frac{d}{d x} (- 13 x^{5})$

Multiply $6$ by $6$ .

$f' (x) = 36 x^{5} + \frac{d}{d x} (- 13 x^{5})$

Evaluate $\frac{d}{d x} [- 13 x^{5}]$ .

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Since $- 13$ is constant with respect to $x$ , the derivative of $- 13 x^{5}$ with respect to $x$ is $- 13 \frac{d}{d x} [x^{5}]$ .

$f' (x) = 36 x^{5} - 13 \frac{d}{d x} x^{5}$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f' (x) = 36 x^{5} - 13 (5 x^{4})$

Multiply $5$ by $- 13$ .

$f' (x) = 36 x^{5} - 65 x^{4}$

Find the second derivative.

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By the Sum Rule, the derivative of $36 x^{5} - 65 x^{4}$ with respect to $x$ is $\frac{d}{d x} [36 x^{5}] + \frac{d}{d x} [- 65 x^{4}]$ .

$f'' (x) = \frac{d}{d x} (36 x^{5}) + \frac{d}{d x} (- 65 x^{4})$

Evaluate $\frac{d}{d x} [36 x^{5}]$ .

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Since $36$ is constant with respect to $x$ , the derivative of $36 x^{5}$ with respect to $x$ is $36 \frac{d}{d x} [x^{5}]$ .

$f'' (x) = 36 \frac{d}{d x} (x^{5}) + \frac{d}{d x} (- 65 x^{4})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f'' (x) = 36 (5 x^{4}) + \frac{d}{d x} (- 65 x^{4})$

Multiply $5$ by $36$ .

$f'' (x) = 180 x^{4} + \frac{d}{d x} (- 65 x^{4})$

Evaluate $\frac{d}{d x} [- 65 x^{4}]$ .

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Since $- 65$ is constant with respect to $x$ , the derivative of $- 65 x^{4}$ with respect to $x$ is $- 65 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 180 x^{4} - 65 \frac{d}{d x} x^{4}$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 180 x^{4} - 65 (4 x^{3})$

Multiply $4$ by $- 65$ .

$f'' (x) = 180 x^{4} - 260 x^{3}$

The second derivative of $f (x)$ with respect to $x$ is $180 x^{4} - 260 x^{3}$ .

$180 x^{4} - 260 x^{3}$

Step 2

Set the second derivative equal to $0$ then solve the equation $180 x^{4} - 260 x^{3} = 0$ .

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Set the second derivative equal to $0$ .

$180 x^{4} - 260 x^{3} = 0$

Factor $20 x^{3}$ out of $180 x^{4} - 260 x^{3}$ .

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Factor $20 x^{3}$ out of $180 x^{4}$ .

$20 x^{3} (9 x) - 260 x^{3} = 0$

Factor $20 x^{3}$ out of $- 260 x^{3}$ .

$20 x^{3} (9 x) + 20 x^{3} (- 13) = 0$

Factor $20 x^{3}$ out of $20 x^{3} (9 x) + 20 x^{3} (- 13)$ .

$20 x^{3} (9 x - 13) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{3} = 0$

$9 x - 13 = 0$

Set $x^{3}$ equal to $0$ and solve for $x$ .

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Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Solve $x^{3} = 0$ for $x$ .

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Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

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

$x = 0$

Set $9 x - 13$ equal to $0$ and solve for $x$ .

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Set $9 x - 13$ equal to $0$ .

$9 x - 13 = 0$

Solve $9 x - 13 = 0$ for $x$ .

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Add $13$ to both sides of the equation.

$9 x = 13$

Divide each term in $9 x = 13$ by $9$ and simplify.

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Divide each term in $9 x = 13$ by $9$ .

$\frac{9 x}{9} = \frac{13}{9}$

Simplify the left side.

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Cancel the common factor of $9$ .

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Cancel the common factor.

$\frac{9 x}{9} = \frac{13}{9}$

Divide $x$ by $1$ .

$x = \frac{13}{9}$

The final solution is all the values that make $20 x^{3} (9 x - 13) = 0$ true.

$x = 0, \frac{13}{9}$

Step 3

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

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Substitute $0$ in $f (x) = 6 x^{6} - 13 x^{5}$ to find the value of $y$ .

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

$f (0) = 6 {(0)}^{6} - 13 {(0)}^{5}$

Simplify the result.

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Simplify each term.

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

$f (0) = 6 \cdot 0 - 13 {(0)}^{5}$

Multiply $6$ by $0$ .

$f (0) = 0 - 13 {(0)}^{5}$

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

$f (0) = 0 - 13 \cdot 0$

Multiply $- 13$ by $0$ .

$f (0) = 0 + 0$

Add $0$ and $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

The point found by substituting $0$ in $f (x) = 6 x^{6} - 13 x^{5}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Substitute $\frac{13}{9}$ in $f (x) = 6 x^{6} - 13 x^{5}$ to find the value of $y$ .

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Replace the variable $x$ with $\frac{13}{9}$ in the expression.

$f (\frac{13}{9}) = 6 {(\frac{13}{9})}^{6} - 13 {(\frac{13}{9})}^{5}$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{13}{9}$ .

$f (\frac{13}{9}) = 6 (\frac{13^{6}}{9^{6}}) - 13 {(\frac{13}{9})}^{5}$

Raise $13$ to the power of $6$ .

$f (\frac{13}{9}) = 6 (\frac{4826809}{9^{6}}) - 13 {(\frac{13}{9})}^{5}$

Raise $9$ to the power of $6$ .

$f (\frac{13}{9}) = 6 (\frac{4826809}{531441}) - 13 {(\frac{13}{9})}^{5}$

Cancel the common factor of $3$ .

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Factor $3$ out of $6$ .

$f (\frac{13}{9}) = 3 (2) (\frac{4826809}{531441}) - 13 {(\frac{13}{9})}^{5}$

Factor $3$ out of $531441$ .

$f (\frac{13}{9}) = 3 \cdot (2 (\frac{4826809}{3 \cdot 177147})) - 13 {(\frac{13}{9})}^{5}$

Cancel the common factor.

$f (\frac{13}{9}) = 3 \cdot (2 (\frac{4826809}{3 \cdot 177147})) - 13 {(\frac{13}{9})}^{5}$

Rewrite the expression.

$f (\frac{13}{9}) = 2 (\frac{4826809}{177147}) - 13 {(\frac{13}{9})}^{5}$

Combine $2$ and $\frac{4826809}{177147}$ .

$f (\frac{13}{9}) = \frac{2 \cdot 4826809}{177147} - 13 {(\frac{13}{9})}^{5}$

Multiply $2$ by $4826809$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - 13 {(\frac{13}{9})}^{5}$

Apply the product rule to $\frac{13}{9}$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - 13 \frac{13^{5}}{9^{5}}$

Raise $13$ to the power of $5$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - 13 \frac{371293}{9^{5}}$

Raise $9$ to the power of $5$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - 13 (\frac{371293}{59049})$

Multiply $- 13 (\frac{371293}{59049})$ .

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Combine $- 13$ and $\frac{371293}{59049}$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} + \frac{- 13 \cdot 371293}{59049}$

Multiply $- 13$ by $371293$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} + \frac{- 4826809}{59049}$

Move the negative in front of the fraction.

$f (\frac{13}{9}) = \frac{9653618}{177147} - \frac{4826809}{59049}$

To write $- \frac{4826809}{59049}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - \frac{4826809}{59049} \cdot \frac{3}{3}$

Write each expression with a common denominator of $177147$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{4826809}{59049}$ by $\frac{3}{3}$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - \frac{4826809 \cdot 3}{59049 \cdot 3}$

Multiply $59049$ by $3$ .

$f (\frac{13}{9}) = \frac{9653618}{177147} - \frac{4826809 \cdot 3}{177147}$

Combine the numerators over the common denominator.

$f (\frac{13}{9}) = \frac{9653618 - 4826809 \cdot 3}{177147}$

Simplify the numerator.

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Multiply $- 4826809$ by $3$ .

$f (\frac{13}{9}) = \frac{9653618 - 14480427}{177147}$

Subtract $14480427$ from $9653618$ .

$f (\frac{13}{9}) = \frac{- 4826809}{177147}$

Move the negative in front of the fraction.

$f (\frac{13}{9}) = - \frac{4826809}{177147}$

The final answer is $- \frac{4826809}{177147}$ .

$- \frac{4826809}{177147}$

The point found by substituting $\frac{13}{9}$ in $f (x) = 6 x^{6} - 13 x^{5}$ is $(\frac{13}{9}, - \frac{4826809}{177147})$ . This point can be an inflection point.

$(\frac{13}{9}, - \frac{4826809}{177147})$

Determine the points that could be inflection points.

$(0, 0), (\frac{13}{9}, - \frac{4826809}{177147})$

Step 4

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

$(- \infty, 0) \cup (0, \frac{13}{9}) \cup (\frac{13}{9}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = 180 {(- 0.1)}^{4} - 260 {(- 0.1)}^{3}$

Simplify the result.

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Simplify each term.

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Raise $- 0.1$ to the power of $4$ .

$f'' (- 0.1) = 180 \cdot 0.0001 - 260 {(- 0.1)}^{3}$

Multiply $180$ by $0.0001$ .

$f'' (- 0.1) = 0.018 - 260 {(- 0.1)}^{3}$

Raise $- 0.1$ to the power of $3$ .

$f'' (- 0.1) = 0.018 - 260 \cdot - 0.001$

Multiply $- 260$ by $- 0.001$ .

$f'' (- 0.1) = 0.018 + 0.26$

Add $0.018$ and $0.26$ .

$f'' (- 0.1) = 0.278$

The final answer is $0.278$ .

$0.278$

At $- 0.1$ , the second derivative is $0.278$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(0, \frac{13}{9})$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $0.7\overline{2}$ in the expression.

$f'' (0.7\overline{2}) = 180 {(0.7\overline{2})}^{4} - 260 {(0.7\overline{2})}^{3}$

Simplify the result.

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Simplify each term.

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Raise $0.7\overline{2}$ to the power of $4$ .

$f'' (0.7\overline{2}) = 180 \cdot 0.27207171 - 260 {(0.7\overline{2})}^{3}$

Multiply $180$ by $0.27207171$ .

$f'' (0.7\overline{2}) = 48.97290809 - 260 {(0.7\overline{2})}^{3}$

Raise $0.7\overline{2}$ to the power of $3$ .

$f'' (0.7\overline{2}) = 48.97290809 - 260 \cdot 0.37671467$

Multiply $- 260$ by $0.37671467$ .

$f'' (0.7\overline{2}) = 48.97290809 - 97.94581618$

Subtract $97.94581618$ from $48.97290809$ .

$f'' (0.7\overline{2}) = - 48.97290809$

The final answer is $- 48.97290809$ .

$- 48.97290809$

At $0.7\overline{2}$ , the second derivative is $- 48.97290809$ . Since this is negative, the second derivative is decreasing on the interval $(0, \frac{13}{9})$

Decreasing on $(0, \frac{13}{9})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{13}{9}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $1.5\overline{4}$ in the expression.

$f'' (1.5\overline{4}) = 180 {(1.5\overline{4})}^{4} - 260 {(1.5\overline{4})}^{3}$

Simplify the result.

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Simplify each term.

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Raise $1.5\overline{4}$ to the power of $4$ .

$f'' (1.5\overline{4}) = 180 \cdot 5.68969731 - 260 {(1.5\overline{4})}^{3}$

Multiply $180$ by $5.68969731$ .

$f'' (1.5\overline{4}) = 1024.14551714 - 260 {(1.5\overline{4})}^{3}$

Raise $1.5\overline{4}$ to the power of $3$ .

$f'' (1.5\overline{4}) = 1024.14551714 - 260 \cdot 3.68397668$

Multiply $- 260$ by $3.68397668$ .

$f'' (1.5\overline{4}) = 1024.14551714 - 957.83393689$

Subtract $957.83393689$ from $1024.14551714$ .

$f'' (1.5\overline{4}) = 66.31158024$

The final answer is $66.31158024$ .

$66.31158024$

At $1.5\overline{4}$ , the second derivative is $66.31158024$ . Since this is positive, the second derivative is increasing on the interval $(\frac{13}{9}, \infty)$ .

Increasing on $(\frac{13}{9}, \infty)$ since $f'' (x) > 0$

Step 8

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 points in this case are $(0, 0), (\frac{13}{9}, - \frac{4826809}{177147})$ .

$(0, 0), (\frac{13}{9}, - \frac{4826809}{177147})$

Step 9