Calculus Examples

$y = x^{3} - 3 x^{2} + 1$

Step 1

Find the first derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $x^{3} - 3 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [1]$

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

$3 x^{2} + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [1]$

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

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

$3 x^{2} - 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

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

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

Multiply $2$ by $- 3$ .

$3 x^{2} - 6 x + \frac{d}{d x} [1]$

Differentiate using the Constant Rule.

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

$3 x^{2} - 6 x + 0$

Add $3 x^{2} - 6 x$ and $0$ .

$3 x^{2} - 6 x$

Step 2

Find the second derivative of the function.

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By the Sum Rule, the derivative of $3 x^{2} - 6 x$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x]$ .

$f'' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x)$

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

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

$f'' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 6 x)$

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

$f'' (x) = 3 (2 x) + \frac{d}{d x} (- 6 x)$

Multiply $2$ by $3$ .

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

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

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

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

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

$f'' (x) = 6 x - 6 \cdot 1$

Multiply $- 6$ by $1$ .

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

Step 3

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

$3 x^{2} - 6 x = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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Differentiate.

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By the Sum Rule, the derivative of $x^{3} - 3 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [1]$ .

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

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

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

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

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

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

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

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

Multiply $2$ by $- 3$ .

$f' (x) = 3 x^{2} - 6 x + \frac{d}{d x} (1)$

Differentiate using the Constant Rule.

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

$f' (x) = 3 x^{2} - 6 x + 0$

Add $3 x^{2} - 6 x$ and $0$ .

$f' (x) = 3 x^{2} - 6 x$

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 6 x$ .

$3 x^{2} - 6 x$

Step 5

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 6 x = 0$ .

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Set the first derivative equal to $0$ .

$3 x^{2} - 6 x = 0$

Factor $3 x$ out of $3 x^{2} - 6 x$ .

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

$3 x (x) - 6 x = 0$

Factor $3 x$ out of $- 6 x$ .

$3 x (x) + 3 x (- 2) = 0$

Factor $3 x$ out of $3 x (x) + 3 x (- 2)$ .

$3 x (x - 2) = 0$

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

$x = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $3 x (x - 2) = 0$ true.

$x = 0, 2$

Step 6

Find the values where the derivative is undefined.

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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 7

Critical points to evaluate.

$x = 0, 2$

Step 8

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.

$6 (0) - 6$

Step 9

Evaluate the second derivative.

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Multiply $6$ by $0$ .

$0 - 6$

Subtract $6$ from $0$ .

$- 6$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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

$f (0) = {(0)}^{3} - 3 {(0)}^{2} + 1$

Simplify the result.

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Simplify each term.

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

$f (0) = 0 - 3 {(0)}^{2} + 1$

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

$f (0) = 0 - 3 \cdot 0 + 1$

Multiply $- 3$ by $0$ .

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

Simplify by adding numbers.

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Add $0$ and $0$ .

$f (0) = 0 + 1$

Add $0$ and $1$ .

$f (0) = 1$

The final answer is $1$ .

$y = 1$

Step 12

Evaluate the second derivative at $x = 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (2) - 6$

Step 13

Evaluate the second derivative.

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Multiply $6$ by $2$ .

$12 - 6$

Subtract $6$ from $12$ .

$6$

Step 14

$x = 2$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 2$ is a local minimum

Step 15

Find the y-value when $x = 2$ .

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

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

Simplify the result.

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Simplify each term.

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

$f (2) = 8 - 3 {(2)}^{2} + 1$

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

$f (2) = 8 - 3 \cdot 4 + 1$

Multiply $- 3$ by $4$ .

$f (2) = 8 - 12 + 1$

Simplify by adding and subtracting.

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Subtract $12$ from $8$ .

$f (2) = - 4 + 1$

Add $- 4$ and $1$ .

$f (2) = - 3$

The final answer is $- 3$ .

$y = - 3$

Step 16

These are the local extrema for $f (x) = x^{3} - 3 x^{2} + 1$ .

$(0, 1)$ is a local maxima

$(2, - 3)$ is a local minima

Step 17