Calculus Examples

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

Find the first derivative of the function.

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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 $1$ .

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

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 $0$ .

$2 x - 3$

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 $1$ .

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

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 $0$ .

$f'' (x) = 2$

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

$2 x - 3 = 0$

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term by $2$ and simplify.

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

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

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

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.

$2$

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

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