Calculus Examples

$f (x) = x^{2} + 5 x - 3$

Step 1

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 $x^{2} + 5 x - 3$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [- 3]$ .

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

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

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

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

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

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

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

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

Multiply $5$ by $1$ .

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

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 $0$ .

$f' (x) = 2 x + 5 + 0$

Add $2 x + 5$ and $0$ .

$f' (x) = 2 x + 5$

The first derivative of $f (x)$ with respect to $x$ is $2 x + 5$ .

$2 x + 5$

Step 2

Set the first derivative equal to $0$ then solve the equation $2 x + 5 = 0$ .

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Set the first derivative equal to $0$ .

$2 x + 5 = 0$

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

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Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 3

The values which make the derivative equal to $0$ are $- \frac{5}{2}$ .

$- \frac{5}{2}$

Step 4

After finding the point that makes the derivative $f' (x) = 2 x + 5$ equal to $0$ or undefined, the interval to check where $f (x) = x^{2} + 5 x - 3$ is increasing and where it is decreasing is $(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \infty)$ .

$(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \frac{5}{2})$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $- \frac{7}{2}$ in the expression.

$f' (- \frac{7}{2}) = 2 (- \frac{7}{2}) + 5$

Simplify the result.

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Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{7}{2}$ into the numerator.

$f' (- \frac{7}{2}) = 2 (\frac{- 7}{2}) + 5$

Cancel the common factor.

$f' (- \frac{7}{2}) = 2 (\frac{- 7}{2}) + 5$

Rewrite the expression.

$f' (- \frac{7}{2}) = - 7 + 5$

Add $- 7$ and $5$ .

$f' (- \frac{7}{2}) = - 2$

The final answer is $- 2$ .

$- 2$

At $x = - \frac{7}{2}$ the derivative is $- 2$ . Since this is negative, the function is decreasing on $(- \infty, - \frac{5}{2})$ .

Decreasing on $(- \infty, - \frac{5}{2})$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- \frac{5}{2}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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

$f' (- \frac{3}{2}) = 2 (- \frac{3}{2}) + 5$

Simplify the result.

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Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{3}{2}$ into the numerator.

$f' (- \frac{3}{2}) = 2 (\frac{- 3}{2}) + 5$

Cancel the common factor.

$f' (- \frac{3}{2}) = 2 (\frac{- 3}{2}) + 5$

Rewrite the expression.

$f' (- \frac{3}{2}) = - 3 + 5$

Add $- 3$ and $5$ .

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

The final answer is $2$ .

$2$

At $x = - \frac{3}{2}$ the derivative is $2$ . Since this is positive, the function is increasing on $(- \frac{5}{2}, \infty)$ .

Increasing on $(- \frac{5}{2}, \infty)$ since $f' (x) > 0$

Step 7

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \frac{5}{2}, \infty)$

Decreasing on: $(- \infty, - \frac{5}{2})$

Step 8