Calculus Examples

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

Find the first derivative of the function.

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By the Sum Rule, the derivative of $4 x^{2} - 3 x + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [1]$ .

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

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

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

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

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

Multiply $2$ by $4$ .

$8 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [1]$

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

$8 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$8 x - 3 \cdot 1 + \frac{d}{d x} [1]$

Multiply $- 3$ by $1$ .

$8 x - 3 + \frac{d}{d x} [1]$

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 $0$ .

$8 x - 3 + 0$

Add $8 x - 3$ and $0$ .

$8 x - 3$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $8 x - 3$ with respect to $x$ is $\frac{d}{d x} [8 x] + \frac{d}{d x} [- 3]$ .

$f'' (x) = \frac{d}{d x} (8 x) + \frac{d}{d x} (- 3)$

Evaluate $\frac{d}{d x} [8 x]$ .

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Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f'' (x) = 8 \frac{d}{d x} (x) + \frac{d}{d x} (- 3)$

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

$f'' (x) = 8 \cdot 1 + \frac{d}{d x} (- 3)$

Multiply $8$ by $1$ .

$f'' (x) = 8 + \frac{d}{d x} (- 3)$

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 $0$ .

$f'' (x) = 8 + 0$

Add $8$ and $0$ .

$f'' (x) = 8$

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

$8 x - 3 = 0$

Add $3$ to both sides of the equation.

$8 x = 3$

Divide each term by $8$ and simplify.

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Divide each term in $8 x = 3$ by $8$ .

$\frac{8 x}{8} = \frac{3}{8}$

Cancel the common factor of $8$ .

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Cancel the common factor.

$\frac{8 x}{8} = \frac{3}{8}$

Divide $x$ by $1$ .

$x = \frac{3}{8}$

Evaluate the second derivative at $x = \frac{3}{8}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$8$

$x = \frac{3}{8}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{3}{8}$ is a local minimum

Find the y-value when $x = \frac{3}{8}$ .

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Replace the variable $x$ with $\frac{3}{8}$ in the expression.

$f (\frac{3}{8}) = 4 {(\frac{3}{8})}^{2} - 3 (\frac{3}{8}) + 1$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{3}{8}$ .

$f (\frac{3}{8}) = 4 (\frac{3^{2}}{8^{2}}) - 3 (\frac{3}{8}) + 1$

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

$f (\frac{3}{8}) = 4 (\frac{9}{8^{2}}) - 3 (\frac{3}{8}) + 1$

Raise $8$ to the power of $2$ .

$f (\frac{3}{8}) = 4 (\frac{9}{64}) - 3 (\frac{3}{8}) + 1$

Cancel the common factor of $4$ .

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Factor $4$ out of $64$ .

$f (\frac{3}{8}) = 4 (\frac{9}{4 (16)}) - 3 (\frac{3}{8}) + 1$

Cancel the common factor.

$f (\frac{3}{8}) = 4 (\frac{9}{4 \cdot 16}) - 3 (\frac{3}{8}) + 1$

Rewrite the expression.

$f (\frac{3}{8}) = \frac{9}{16} - 3 (\frac{3}{8}) + 1$

Multiply $- 3 (\frac{3}{8})$ .

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Combine $- 3$ and $\frac{3}{8}$ .

$f (\frac{3}{8}) = \frac{9}{16} + \frac{- 3 \cdot 3}{8} + 1$

Multiply $- 3$ by $3$ .

$f (\frac{3}{8}) = \frac{9}{16} + \frac{- 9}{8} + 1$

Move the negative in front of the fraction.

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9}{8} + 1$

Find the common denominator.

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Multiply $\frac{9}{8}$ by $\frac{2}{2}$ .

$f (\frac{3}{8}) = \frac{9}{16} - (\frac{9}{8} \cdot \frac{2}{2}) + 1$

Multiply $\frac{9}{8}$ and $\frac{2}{2}$ .

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9 \cdot 2}{8 \cdot 2} + 1$

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

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9 \cdot 2}{8 \cdot 2} + \frac{1}{1}$

Multiply $\frac{1}{1}$ by $\frac{16}{16}$ .

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9 \cdot 2}{8 \cdot 2} + \frac{1}{1} \cdot \frac{16}{16}$

Multiply $\frac{1}{1}$ and $\frac{16}{16}$ .

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9 \cdot 2}{8 \cdot 2} + \frac{16}{16}$

Reorder the factors of $8 \cdot 2$ .

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9 \cdot 2}{2 \cdot 8} + \frac{16}{16}$

Multiply $2$ by $8$ .

$f (\frac{3}{8}) = \frac{9}{16} - \frac{9 \cdot 2}{16} + \frac{16}{16}$

Combine fractions.

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Combine fractions with similar denominators.

$f (\frac{3}{8}) = \frac{9 - 9 \cdot 2 + 16}{16}$

Multiply $- 9$ by $2$ .

$f (\frac{3}{8}) = \frac{9 - 18 + 16}{16}$

Simplify the numerator.

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Subtract $18$ from $9$ .

$f (\frac{3}{8}) = \frac{- 9 + 16}{16}$

Add $- 9$ and $16$ .

$f (\frac{3}{8}) = \frac{7}{16}$

The final answer is $\frac{7}{16}$ .

$y = \frac{7}{16}$

These are the local extrema for $f (x) = 4 x^{2} - 3 x + 1$ .

$(\frac{3}{8}, \frac{7}{16})$ is a local minima

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