Calculus Examples

$f (x) = 3 x^{5} - 10 x^{4} + 1$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $3 x^{5} - 10 x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [3 x^{5}] + \frac{d}{d x} [- 10 x^{4}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [3 x^{5}] + \frac{d}{d x} [- 10 x^{4}] + \frac{d}{d x} [1]$

Step 1.1.2

Evaluate $\frac{d}{d x} [3 x^{5}]$ .

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Step 1.1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{5}$ with respect to $x$ is $3 \frac{d}{d x} [x^{5}]$ .

$3 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 10 x^{4}] + \frac{d}{d x} [1]$

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 $n = 5$ .

$3 (5 x^{4}) + \frac{d}{d x} [- 10 x^{4}] + \frac{d}{d x} [1]$

Step 1.1.2.3

Multiply $5$ by $3$ .

$15 x^{4} + \frac{d}{d x} [- 10 x^{4}] + \frac{d}{d x} [1]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 10 x^{4}]$ .

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Step 1.1.3.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x^{4}$ with respect to $x$ is $- 10 \frac{d}{d x} [x^{4}]$ .

$15 x^{4} - 10 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$

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 $n = 4$ .

$15 x^{4} - 10 (4 x^{3}) + \frac{d}{d x} [1]$

Step 1.1.3.3

Multiply $4$ by $- 10$ .

$15 x^{4} - 40 x^{3} + \frac{d}{d x} [1]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$15 x^{4} - 40 x^{3} + 0$

Step 1.1.4.2

Add $15 x^{4} - 40 x^{3}$ and $0$ .

$f' (x) = 15 x^{4} - 40 x^{3}$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $15 x^{4} - 40 x^{3}$ with respect to $x$ is $\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [- 40 x^{3}]$ .

$\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [- 40 x^{3}]$

Step 1.2.2

Evaluate $\frac{d}{d x} [15 x^{4}]$ .

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Step 1.2.2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{4}$ with respect to $x$ is $15 \frac{d}{d x} [x^{4}]$ .

$15 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 40 x^{3}]$

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 $n = 4$ .

$15 (4 x^{3}) + \frac{d}{d x} [- 40 x^{3}]$

Step 1.2.2.3

Multiply $4$ by $15$ .

$60 x^{3} + \frac{d}{d x} [- 40 x^{3}]$

Step 1.2.3

Evaluate $\frac{d}{d x} [- 40 x^{3}]$ .

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Step 1.2.3.1

Since $- 40$ is constant with respect to $x$ , the derivative of $- 40 x^{3}$ with respect to $x$ is $- 40 \frac{d}{d x} [x^{3}]$ .

$60 x^{3} - 40 \frac{d}{d x} [x^{3}]$

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 $n = 3$ .

$60 x^{3} - 40 (3 x^{2})$

Step 1.2.3.3

Multiply $3$ by $- 40$ .

$f'' (x) = 60 x^{3} - 120 x^{2}$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $60 x^{3} - 120 x^{2}$ .

$60 x^{3} - 120 x^{2}$

Step 2

Set the second derivative equal to $0$ then solve the equation $60 x^{3} - 120 x^{2} = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$60 x^{3} - 120 x^{2} = 0$

Step 2.2

Factor $60 x^{2}$ out of $60 x^{3} - 120 x^{2}$ .

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Step 2.2.1

Factor $60 x^{2}$ out of $60 x^{3}$ .

$60 x^{2} (x) - 120 x^{2} = 0$

Step 2.2.2

Factor $60 x^{2}$ out of $- 120 x^{2}$ .

$60 x^{2} (x) + 60 x^{2} (- 2) = 0$

Step 2.2.3

Factor $60 x^{2}$ out of $60 x^{2} (x) + 60 x^{2} (- 2)$ .

$60 x^{2} (x - 2) = 0$

Step 2.3

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

$x^{2} = 0$

$x - 2 = 0$

Step 2.4

Set $x^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Step 2.4.2

Solve $x^{2} = 0$ for $x$ .

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Step 2.4.2.1

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

$x = \pm \sqrt{0}$

Step 2.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.4.2.2.1

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

$x = \pm \sqrt{0^{2}}$

Step 2.4.2.2.2

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

$x = \pm 0$

Step 2.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.5

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

The final solution is all the values that make $60 x^{2} (x - 2) = 0$ true.

$x = 0, 2$

Step 3

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

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Step 3.1

Substitute $0$ in $f (x) = 3 x^{5} - 10 x^{4} + 1$ to find the value of $y$ .

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Step 3.1.1

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

$f (0) = 3 {(0)}^{5} - 10 {(0)}^{4} + 1$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

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

$f (0) = 3 \cdot 0 - 10 {(0)}^{4} + 1$

Step 3.1.2.1.2

Multiply $3$ by $0$ .

$f (0) = 0 - 10 {(0)}^{4} + 1$

Step 3.1.2.1.3

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

$f (0) = 0 - 10 \cdot 0 + 1$

Step 3.1.2.1.4

Multiply $- 10$ by $0$ .

$f (0) = 0 + 0 + 1$

Step 3.1.2.2

Simplify by adding numbers.

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Step 3.1.2.2.1

Add $0$ and $0$ .

$f (0) = 0 + 1$

Step 3.1.2.2.2

Add $0$ and $1$ .

$f (0) = 1$

Step 3.1.2.3

The final answer is $1$ .

$1$

Step 3.2

The point found by substituting $0$ in $f (x) = 3 x^{5} - 10 x^{4} + 1$ is $(0, 1)$ . This point can be an inflection point.

$(0, 1)$

Step 3.3

Substitute $2$ in $f (x) = 3 x^{5} - 10 x^{4} + 1$ to find the value of $y$ .

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Step 3.3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = 3 {(2)}^{5} - 10 {(2)}^{4} + 1$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Raise $2$ to the power of $5$ .

$f (2) = 3 \cdot 32 - 10 {(2)}^{4} + 1$

Step 3.3.2.1.2

Multiply $3$ by $32$ .

$f (2) = 96 - 10 {(2)}^{4} + 1$

Step 3.3.2.1.3

Raise $2$ to the power of $4$ .

$f (2) = 96 - 10 \cdot 16 + 1$

Step 3.3.2.1.4

Multiply $- 10$ by $16$ .

$f (2) = 96 - 160 + 1$

Step 3.3.2.2

Simplify by adding and subtracting.

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Step 3.3.2.2.1

Subtract $160$ from $96$ .

$f (2) = - 64 + 1$

Step 3.3.2.2.2

Add $- 64$ and $1$ .

$f (2) = - 63$

Step 3.3.2.3

The final answer is $- 63$ .

$- 63$

Step 3.4

The point found by substituting $2$ in $f (x) = 3 x^{5} - 10 x^{4} + 1$ is $(2, - 63)$ . This point can be an inflection point.

$(2, - 63)$

Step 3.5

Determine the points that could be inflection points.

$(0, 1), (2, - 63)$

Step 4

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

$(- \infty, 0) \cup (0, 2) \cup (2, \infty)$

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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Step 5.1

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

$f'' (- 0.1) = 60 {(- 0.1)}^{3} - 120 {(- 0.1)}^{2}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

$f'' (- 0.1) = 60 \cdot - 0.001 - 120 {(- 0.1)}^{2}$

Step 5.2.1.2

Multiply $60$ by $- 0.001$ .

$f'' (- 0.1) = - 0.06 - 120 {(- 0.1)}^{2}$

Step 5.2.1.3

Raise $- 0.1$ to the power of $2$ .

$f'' (- 0.1) = - 0.06 - 120 \cdot 0.01$

Step 5.2.1.4

Multiply $- 120$ by $0.01$ .

$f'' (- 0.1) = - 0.06 - 1.2$

Step 5.2.2

Subtract $1.2$ from $- 0.06$ .

$f'' (- 0.1) = - 1.26$

Step 5.2.3

The final answer is $- 1.26$ .

$- 1.26$

Step 5.3

At $- 0.1$ , the second derivative is $- 1.26$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

Decreasing on $(- \infty, 0)$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(0, 2)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

$f'' (1) = 60 {(1)}^{3} - 120 {(1)}^{2}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

One to any power is one.

$f'' (1) = 60 \cdot 1 - 120 {(1)}^{2}$

Step 6.2.1.2

Multiply $60$ by $1$ .

$f'' (1) = 60 - 120 {(1)}^{2}$

Step 6.2.1.3

One to any power is one.

$f'' (1) = 60 - 120 \cdot 1$

Step 6.2.1.4

Multiply $- 120$ by $1$ .

$f'' (1) = 60 - 120$

Step 6.2.2

Subtract $120$ from $60$ .

$f'' (1) = - 60$

Step 6.2.3

The final answer is $- 60$ .

$- 60$

Step 6.3

At $1$ , the second derivative is $- 60$ . Since this is negative, the second derivative is decreasing on the interval $(0, 2)$

Decreasing on $(0, 2)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(2, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $2.1$ in the expression.

$f'' (2.1) = 60 {(2.1)}^{3} - 120 {(2.1)}^{2}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $2.1$ to the power of $3$ .

$f'' (2.1) = 60 \cdot 9.261 - 120 {(2.1)}^{2}$

Step 7.2.1.2

Multiply $60$ by $9.261$ .

$f'' (2.1) = 555.66 - 120 {(2.1)}^{2}$

Step 7.2.1.3

Raise $2.1$ to the power of $2$ .

$f'' (2.1) = 555.66 - 120 \cdot 4.41$

Step 7.2.1.4

Multiply $- 120$ by $4.41$ .

$f'' (2.1) = 555.66 - 529.2$

Step 7.2.2

Subtract $529.2$ from $555.66$ .

$f'' (2.1) = 26.46$

Step 7.2.3

The final answer is $26.46$ .

$26.46$

Step 7.3

At $2.1$ , the second derivative is $26.46$ . Since this is positive, the second derivative is increasing on the interval $(2, \infty)$ .

Increasing on $(2, \infty)$ 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 point in this case is $(2, - 63)$ .

$(2, - 63)$

Step 9