Calculus Examples

$y = 3 x^{5} - 5 x^{3} + 1$

Step 1

Write $y = 3 x^{5} - 5 x^{3} + 1$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $5$ by $3$ .

$15 x^{4} + \frac{d}{d x} [- 5 x^{3}] + \frac{d}{d x} [1]$

Step 2.3

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

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Step 2.3.1

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

$15 x^{4} - 5 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [1]$

Step 2.3.2

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

$15 x^{4} - 5 (3 x^{2}) + \frac{d}{d x} [1]$

Step 2.3.3

Multiply $3$ by $- 5$ .

$15 x^{4} - 15 x^{2} + \frac{d}{d x} [1]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$15 x^{4} - 15 x^{2} + 0$

Step 2.4.2

Add $15 x^{4} - 15 x^{2}$ and $0$ .

$15 x^{4} - 15 x^{2}$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $15 x^{4} - 15 x^{2}$ with respect to $x$ is $\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [- 15 x^{2}]$ .

$f'' (x) = \frac{d}{d x} (15 x^{4}) + \frac{d}{d x} (- 15 x^{2})$

Step 3.2

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

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Step 3.2.1

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

$f'' (x) = 15 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 15 x^{2})$

Step 3.2.2

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

$f'' (x) = 15 (4 x^{3}) + \frac{d}{d x} (- 15 x^{2})$

Step 3.2.3

Multiply $4$ by $15$ .

$f'' (x) = 60 x^{3} + \frac{d}{d x} (- 15 x^{2})$

Step 3.3

Evaluate $\frac{d}{d x} [- 15 x^{2}]$ .

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Step 3.3.1

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15 x^{2}$ with respect to $x$ is $- 15 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 60 x^{3} - 15 \frac{d}{d x} x^{2}$

Step 3.3.2

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

$f'' (x) = 60 x^{3} - 15 (2 x)$

Step 3.3.3

Multiply $2$ by $- 15$ .

$f'' (x) = 60 x^{3} - 30 x$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$15 x^{4} - 15 x^{2} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

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

$f' (x) = \frac{d}{d x} (3 x^{5}) + \frac{d}{d x} (- 5 x^{3}) + \frac{d}{d x} (1)$

Step 5.1.2

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

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Step 5.1.2.1

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

$f' (x) = 3 \frac{d}{d x} (x^{5}) + \frac{d}{d x} (- 5 x^{3}) + \frac{d}{d x} (1)$

Step 5.1.2.2

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

$f' (x) = 3 (5 x^{4}) + \frac{d}{d x} (- 5 x^{3}) + \frac{d}{d x} (1)$

Step 5.1.2.3

Multiply $5$ by $3$ .

$f' (x) = 15 x^{4} + \frac{d}{d x} (- 5 x^{3}) + \frac{d}{d x} (1)$

Step 5.1.3

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

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Step 5.1.3.1

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

$f' (x) = 15 x^{4} - 5 \frac{d}{d x} x^{3} + \frac{d}{d x} (1)$

Step 5.1.3.2

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

$f' (x) = 15 x^{4} - 5 (3 x^{2}) + \frac{d}{d x} (1)$

Step 5.1.3.3

Multiply $3$ by $- 5$ .

$f' (x) = 15 x^{4} - 15 x^{2} + \frac{d}{d x} (1)$

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

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

$f' (x) = 15 x^{4} - 15 x^{2} + 0$

Step 5.1.4.2

Add $15 x^{4} - 15 x^{2}$ and $0$ .

$f' (x) = 15 x^{4} - 15 x^{2}$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $15 x^{4} - 15 x^{2}$ .

$15 x^{4} - 15 x^{2}$

Step 6

Set the first derivative equal to $0$ then solve the equation $15 x^{4} - 15 x^{2} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$15 x^{4} - 15 x^{2} = 0$

Step 6.2

Factor the left side of the equation.

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Step 6.2.1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$15 {(x^{2})}^{2} - 15 x^{2} = 0$

Step 6.2.2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$15 u^{2} - 15 u = 0$

Step 6.2.3

Factor $15 u$ out of $15 u^{2} - 15 u$ .

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Step 6.2.3.1

Factor $15 u$ out of $15 u^{2}$ .

$15 u (u) - 15 u = 0$

Step 6.2.3.2

Factor $15 u$ out of $- 15 u$ .

$15 u (u) + 15 u (- 1) = 0$

Step 6.2.3.3

Factor $15 u$ out of $15 u (u) + 15 u (- 1)$ .

$15 u (u - 1) = 0$

Step 6.2.4

Replace all occurrences of $u$ with $x^{2}$ .

$15 x^{2} (x^{2} - 1) = 0$

Step 6.3

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

$x^{2} = 0$

$x^{2} - 1 = 0$

Step 6.4

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

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Step 6.4.1

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

$x^{2} = 0$

Step 6.4.2

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

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Step 6.4.2.1

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

$x = \pm \sqrt{0}$

Step 6.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 6.4.2.2.1

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

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

Step 6.4.2.2.2

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

$x = \pm 0$

Step 6.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 6.5

Set $x^{2} - 1$ equal to $0$ and solve for $x$ .

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Step 6.5.1

Set $x^{2} - 1$ equal to $0$ .

$x^{2} - 1 = 0$

Step 6.5.2

Solve $x^{2} - 1 = 0$ for $x$ .

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Step 6.5.2.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 6.5.2.2

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

$x = \pm \sqrt{1}$

Step 6.5.2.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 6.5.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.5.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 6.5.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 6.5.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 6.6

The final solution is all the values that make $15 x^{2} (x^{2} - 1) = 0$ true.

$x = 0, 1, - 1$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 8

Critical points to evaluate.

$x = 0, 1, - 1$

Step 9

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$60 {(0)}^{3} - 30 \cdot 0$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

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

$60 \cdot 0 - 30 \cdot 0$

Step 10.1.2

Multiply $60$ by $0$ .

$0 - 30 \cdot 0$

Step 10.1.3

Multiply $- 30$ by $0$ .

$0 + 0$

Step 10.2

Add $0$ and $0$ .

$0$

Step 11

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 11.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

Step 11.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - 1)$ in the first derivative $15 x^{4} - 15 x^{2}$ to check if the result is negative or positive.

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Step 11.2.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = 15 {(- 4)}^{4} - 15 {(- 4)}^{2}$

Step 11.2.2

Simplify the result.

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Step 11.2.2.1

Simplify each term.

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Step 11.2.2.1.1

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

$f' (- 4) = 15 \cdot 256 - 15 {(- 4)}^{2}$

Step 11.2.2.1.2

Multiply $15$ by $256$ .

$f' (- 4) = 3840 - 15 {(- 4)}^{2}$

Step 11.2.2.1.3

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

$f' (- 4) = 3840 - 15 \cdot 16$

Step 11.2.2.1.4

Multiply $- 15$ by $16$ .

$f' (- 4) = 3840 - 240$

Step 11.2.2.2

Subtract $240$ from $3840$ .

$f' (- 4) = 3600$

Step 11.2.2.3

The final answer is $3600$ .

$3600$

Step 11.3

Substitute any number, such as $- 0.5$ , from the interval $(- 1, 0)$ in the first derivative $15 x^{4} - 15 x^{2}$ to check if the result is negative or positive.

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Step 11.3.1

Replace the variable $x$ with $- 0.5$ in the expression.

$f' (- 0.5) = 15 {(- 0.5)}^{4} - 15 {(- 0.5)}^{2}$

Step 11.3.2

Simplify the result.

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Step 11.3.2.1

Simplify each term.

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Step 11.3.2.1.1

Raise $- 0.5$ to the power of $4$ .

$f' (- 0.5) = 15 \cdot 0.0625 - 15 {(- 0.5)}^{2}$

Step 11.3.2.1.2

Multiply $15$ by $0.0625$ .

$f' (- 0.5) = 0.9375 - 15 {(- 0.5)}^{2}$

Step 11.3.2.1.3

Raise $- 0.5$ to the power of $2$ .

$f' (- 0.5) = 0.9375 - 15 \cdot 0.25$

Step 11.3.2.1.4

Multiply $- 15$ by $0.25$ .

$f' (- 0.5) = 0.9375 - 3.75$

Step 11.3.2.2

Subtract $3.75$ from $0.9375$ .

$f' (- 0.5) = - 2.8125$

Step 11.3.2.3

The final answer is $- 2.8125$ .

$- 2.8125$

Step 11.4

Substitute any number, such as $0.5$ , from the interval $(0, 1)$ in the first derivative $15 x^{4} - 15 x^{2}$ to check if the result is negative or positive.

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Step 11.4.1

Replace the variable $x$ with $0.5$ in the expression.

$f' (0.5) = 15 {(0.5)}^{4} - 15 {(0.5)}^{2}$

Step 11.4.2

Simplify the result.

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Step 11.4.2.1

Simplify each term.

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Step 11.4.2.1.1

Raise $0.5$ to the power of $4$ .

$f' (0.5) = 15 \cdot 0.0625 - 15 {(0.5)}^{2}$

Step 11.4.2.1.2

Multiply $15$ by $0.0625$ .

$f' (0.5) = 0.9375 - 15 {(0.5)}^{2}$

Step 11.4.2.1.3

Raise $0.5$ to the power of $2$ .

$f' (0.5) = 0.9375 - 15 \cdot 0.25$

Step 11.4.2.1.4

Multiply $- 15$ by $0.25$ .

$f' (0.5) = 0.9375 - 3.75$

Step 11.4.2.2

Subtract $3.75$ from $0.9375$ .

$f' (0.5) = - 2.8125$

Step 11.4.2.3

The final answer is $- 2.8125$ .

$- 2.8125$

Step 11.5

Substitute any number, such as $4$ , from the interval $(1, \infty)$ in the first derivative $15 x^{4} - 15 x^{2}$ to check if the result is negative or positive.

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Step 11.5.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = 15 {(4)}^{4} - 15 {(4)}^{2}$

Step 11.5.2

Simplify the result.

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Step 11.5.2.1

Simplify each term.

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Step 11.5.2.1.1

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

$f' (4) = 15 \cdot 256 - 15 {(4)}^{2}$

Step 11.5.2.1.2

Multiply $15$ by $256$ .

$f' (4) = 3840 - 15 {(4)}^{2}$

Step 11.5.2.1.3

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

$f' (4) = 3840 - 15 \cdot 16$

Step 11.5.2.1.4

Multiply $- 15$ by $16$ .

$f' (4) = 3840 - 240$

Step 11.5.2.2

Subtract $240$ from $3840$ .

$f' (4) = 3600$

Step 11.5.2.3

The final answer is $3600$ .

$3600$

Step 11.6

Since the first derivative changed signs from positive to negative around $x = - 1$ , then $x = - 1$ is a local maximum.

$x = - 1$ is a local maximum

Step 11.7

Since the first derivative did not change signs around $x = 0$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 11.8

Since the first derivative changed signs from negative to positive around $x = 1$ , then $x = 1$ is a local minimum.

$x = 1$ is a local minimum

Step 11.9

These are the local extrema for $f (x) = 3 x^{5} - 5 x^{3} + 1$ .

$x = - 1$ is a local maximum

$x = 1$ is a local minimum

$x = - 1$ is a local maximum

$x = 1$ is a local minimum

Step 12