Calculus Examples

$f (x) = 2.9 + 3.8 x - 0.9 x^{2}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $2.9 + 3.8 x - 0.9 x^{2}$ with respect to $x$ is $\frac{d}{d x} [2.9] + \frac{d}{d x} [3.8 x] + \frac{d}{d x} [- 0.9 x^{2}]$ .

$\frac{d}{d x} [2.9] + \frac{d}{d x} [3.8 x] + \frac{d}{d x} [- 0.9 x^{2}]$

Step 1.1.1.2

Since $2.9$ is constant with respect to $x$ , the derivative of $2.9$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [3.8 x] + \frac{d}{d x} [- 0.9 x^{2}]$

Step 1.1.2

Evaluate $\frac{d}{d x} [3.8 x]$ .

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Step 1.1.2.1

Since $3.8$ is constant with respect to $x$ , the derivative of $3.8 x$ with respect to $x$ is $3.8 \frac{d}{d x} [x]$ .

$0 + 3.8 \frac{d}{d x} [x] + \frac{d}{d x} [- 0.9 x^{2}]$

Step 1.1.2.2

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

$0 + 3.8 \cdot 1 + \frac{d}{d x} [- 0.9 x^{2}]$

Step 1.1.2.3

Multiply $3.8$ by $1$ .

$0 + 3.8 + \frac{d}{d x} [- 0.9 x^{2}]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 0.9 x^{2}]$ .

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Step 1.1.3.1

Since $- 0.9$ is constant with respect to $x$ , the derivative of $- 0.9 x^{2}$ with respect to $x$ is $- 0.9 \frac{d}{d x} [x^{2}]$ .

$0 + 3.8 - 0.9 \frac{d}{d x} [x^{2}]$

Step 1.1.3.2

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

$0 + 3.8 - 0.9 (2 x)$

Step 1.1.3.3

Multiply $2$ by $- 0.9$ .

$0 + 3.8 - 1.8 x$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Add $0$ and $3.8$ .

$3.8 - 1.8 x$

Step 1.1.4.2

Reorder terms.

$f' (x) = - 1.8 x + 3.8$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 1.8 x + 3.8$ .

$- 1.8 x + 3.8$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 1.8 x + 3.8 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 1.8 x + 3.8 = 0$

Step 2.2

Subtract $3.8$ from both sides of the equation.

$- 1.8 x = - 3.8$

Step 2.3

Divide each term in $- 1.8 x = - 3.8$ by $- 1.8$ and simplify.

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Step 2.3.1

Divide each term in $- 1.8 x = - 3.8$ by $- 1.8$ .

$\frac{- 1.8 x}{- 1.8} = \frac{- 3.8}{- 1.8}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $- 1.8$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{- 1.8 x}{- 1.8} = \frac{- 3.8}{- 1.8}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 3.8}{- 1.8}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 3.8$ by $- 1.8$ .

$x = 2.\overline{1}$

Step 3

The values which make the derivative equal to $0$ are $2.\overline{1}$ .

$2.\overline{1}$

Step 4

After finding the point that makes the derivative $f' (x) = - 1.8 x + 3.8$ equal to $0$ or undefined, the interval to check where $f (x) = 2.9 + 3.8 x - 0.9 x^{2}$ is increasing and where it is decreasing is $(- \infty, 2.\overline{1}) \cup (2.\overline{1}, \infty)$ .

$(- \infty, 2.\overline{1}) \cup (2.\overline{1}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 2.\overline{1})$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $1.\overline{1}$ in the expression.

$f' (1.\overline{1}) = - 1.8 \cdot 1.\overline{1} + 3.8$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $- 1.8$ by $1.\overline{1}$ .

$f' (1.\overline{1}) = - 2 + 3.8$

Step 5.2.2

Add $- 2$ and $3.8$ .

$f' (1.\overline{1}) = 1.8$

Step 5.2.3

The final answer is $1.8$ .

$1.8$

Step 5.3

At $x = 1.\overline{1}$ the derivative is $1.8$ . Since this is positive, the function is increasing on $(- \infty, 2.\overline{1})$ .

Increasing on $(- \infty, 2.\overline{1})$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(2.\overline{1}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $3.\overline{1}$ in the expression.

$f' (3.\overline{1}) = - 1.8 \cdot 3.\overline{1} + 3.8$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $- 1.8$ by $3.\overline{1}$ .

$f' (3.\overline{1}) = - 5.6 + 3.8$

Step 6.2.2

Add $- 5.6$ and $3.8$ .

$f' (3.\overline{1}) = - 1.8$

Step 6.2.3

The final answer is $- 1.8$ .

$- 1.8$

Step 6.3

At $x = 3.\overline{1}$ the derivative is $- 1.8$ . Since this is negative, the function is decreasing on $(2.\overline{1}, \infty)$ .

Decreasing on $(2.\overline{1}, \infty)$ since $f' (x) < 0$

Step 7

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

Increasing on: $(- \infty, 2.\overline{1})$

Decreasing on: $(2.\overline{1}, \infty)$

Step 8