Calculus Examples

$f (x) = x^{2} - 3 x + 4$

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 $x^{2} - 3 x + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

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

$2 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$

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

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

$2 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [4]$

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

$2 x - 3 \cdot 1 + \frac{d}{d x} [4]$

Multiply $- 3$ by $1$ .

$2 x - 3 + \frac{d}{d x} [4]$

Differentiate using the Constant Rule.

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Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$2 x - 3 + 0$

Add $2 x - 3$ and $0$ .

$2 x - 3$

Step 2

Find the second derivative of the function.

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Differentiate using the Constant Rule.

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Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f'' (x) = 2 + 0$

Add $2$ and $0$ .

$f'' (x) = 2$

Step 3

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

$2 x - 3 = 0$

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 $x^{2} - 3 x + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [4]$ .

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

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

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

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

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

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

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

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

Multiply $- 3$ by $1$ .

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

Differentiate using the Constant Rule.

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Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

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

Add $2 x - 3$ and $0$ .

$f' (x) = 2 x - 3$

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

$2 x - 3$

Step 5

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

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Set the first derivative equal to $0$ .

$2 x - 3 = 0$

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

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Divide each term in $2 x = 3$ by $2$ .

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

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

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.

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

Step 8

Evaluate the second derivative at $x = \frac{3}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2$

Step 9

$x = \frac{3}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{3}{2}$ is a local minimum

Step 10

Find the y-value when $x = \frac{3}{2}$ .

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Replace the variable $x$ with $\frac{3}{2}$ in the expression.

$f (\frac{3}{2}) = {(\frac{3}{2})}^{2} - 3 (\frac{3}{2}) + 4$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{3}{2}$ .

$f (\frac{3}{2}) = \frac{3^{2}}{2^{2}} - 3 (\frac{3}{2}) + 4$

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

$f (\frac{3}{2}) = \frac{9}{2^{2}} - 3 (\frac{3}{2}) + 4$

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

$f (\frac{3}{2}) = \frac{9}{4} - 3 (\frac{3}{2}) + 4$

Multiply $- 3 (\frac{3}{2})$ .

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Combine $- 3$ and $\frac{3}{2}$ .

$f (\frac{3}{2}) = \frac{9}{4} + \frac{- 3 \cdot 3}{2} + 4$

Multiply $- 3$ by $3$ .

$f (\frac{3}{2}) = \frac{9}{4} + \frac{- 9}{2} + 4$

Move the negative in front of the fraction.

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9}{2} + 4$

Find the common denominator.

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Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (\frac{3}{2}) = \frac{9}{4} - (\frac{9}{2} \cdot \frac{2}{2}) + 4$

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{2 \cdot 2} + 4$

Write $4$ as a fraction with denominator $1$ .

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{2 \cdot 2} + \frac{4}{1}$

Multiply $\frac{4}{1}$ by $\frac{4}{4}$ .

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{2 \cdot 2} + \frac{4}{1} \cdot \frac{4}{4}$

Multiply $\frac{4}{1}$ by $\frac{4}{4}$ .

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{2 \cdot 2} + \frac{4 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{4} + \frac{4 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (\frac{3}{2}) = \frac{9 - 9 \cdot 2 + 4 \cdot 4}{4}$

Simplify each term.

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Multiply $- 9$ by $2$ .

$f (\frac{3}{2}) = \frac{9 - 18 + 4 \cdot 4}{4}$

Multiply $4$ by $4$ .

$f (\frac{3}{2}) = \frac{9 - 18 + 16}{4}$

Simplify by adding and subtracting.

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Subtract $18$ from $9$ .

$f (\frac{3}{2}) = \frac{- 9 + 16}{4}$

Add $- 9$ and $16$ .

$f (\frac{3}{2}) = \frac{7}{4}$

The final answer is $\frac{7}{4}$ .

$y = \frac{7}{4}$

Step 11

These are the local extrema for $f (x) = x^{2} - 3 x + 4$ .

$(\frac{3}{2}, \frac{7}{4})$ is a local minima

Step 12

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