Calculus Examples

$f (x) = 3 x^{2} + 6 x - 5$ , $[- 5, 1]$

If $f$ is continuous on the interval $[a, b]$ and differentiable on $(a, b)$ , then at least one real number $c$ exists in the interval $(a, b)$ such that $f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at $x = c$ and the slope of the line through the points $(a, f (a))$ and $(b, f (b))$ .

If $f (x)$ is continuous on $[a, b]$

and if $f (x)$ differentiable on $(a, b)$ ,

then there exists at least one point, $c$ in $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Check if $f (x) = 3 x^{2} + 6 x - 5$ is continuous.

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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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

$f (x)$ is continuous on $[- 5, 1]$ .

The function is continuous.

Find the derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $3 x^{2} + 6 x - 5$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 5]$ .

$f' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 5)$

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 5)$

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

$f' (x) = 3 (2 x) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 5)$

Multiply $2$ by $3$ .

$f' (x) = 6 x + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 5)$

Evaluate $\frac{d}{d x} [6 x]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$f' (x) = 6 x + 6 \frac{d}{d x} (x) + \frac{d}{d x} (- 5)$

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

$f' (x) = 6 x + 6 \cdot 1 + \frac{d}{d x} (- 5)$

Multiply $6$ by $1$ .

$f' (x) = 6 x + 6 + \frac{d}{d x} (- 5)$

Differentiate using the Constant Rule.

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Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f' (x) = 6 x + 6 + 0$

Add $6 x + 6$ and $0$ .

$f' (x) = 6 x + 6$

The first derivative of $f (x)$ with respect to $x$ is $6 x + 6$ .

$6 x + 6$

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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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

$f' (x)$ is continuous on $(- 5, 1)$ .

The function is continuous.

The function is differentiable on $(- 5, 1)$ because the derivative is continuous on $(- 5, 1)$ .

The function is differentiable.

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[- 5, 1]$ and differentiable on $(- 5, 1)$ .

$f (x)$ is continuous on $[- 5, 1]$ and differentiable on $(- 5, 1)$ .

Evaluate $f (a)$ from the interval $[- 5, 1]$ .

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Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = 3 {(- 5)}^{2} + 6 (- 5) - 5$

Simplify the result.

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Simplify each term.

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Raise $- 5$ to the power of $2$ .

$f (- 5) = 3 \cdot 25 + 6 (- 5) - 5$

Multiply $3$ by $25$ .

$f (- 5) = 75 + 6 (- 5) - 5$

Multiply $6$ by $- 5$ .

$f (- 5) = 75 - 30 - 5$

Simplify by subtracting numbers.

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Subtract $30$ from $75$ .

$f (- 5) = 45 - 5$

Subtract $5$ from $45$ .

$f (- 5) = 40$

The final answer is $40$ .

$40$

Evaluate $f (b)$ from the interval $[- 5, 1]$ .

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Replace the variable $x$ with $1$ in the expression.

$f (1) = 3 {(1)}^{2} + 6 (1) - 5$

Simplify the result.

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Simplify each term.

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One to any power is one.

$f (1) = 3 \cdot 1 + 6 (1) - 5$

Multiply $3$ by $1$ .

$f (1) = 3 + 6 (1) - 5$

Multiply $6$ by $1$ .

$f (1) = 3 + 6 - 5$

Simplify by adding and subtracting.

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Add $3$ and $6$ .

$f (1) = 9 - 5$

Subtract $5$ from $9$ .

$f (1) = 4$

The final answer is $4$ .

$4$

Solve $6 x + 6 = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $6 x + 6 = \frac{- (f (1) + f (- 5))}{- (1 - 5)}$ .

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Simplify $\frac{(4) - (40)}{(1) - (- 5)}$ .

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Simplify the numerator.

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Multiply $- 1$ by $40$ .

$6 x + 6 = \frac{4 - 40}{(1) - (- 5)}$

Subtract $40$ from $4$ .

$6 x + 6 = \frac{- 36}{(1) - (- 5)}$

Simplify the denominator.

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Multiply $- 1$ by $- 5$ .

$6 x + 6 = \frac{- 36}{1 + 5}$

Add $1$ and $5$ .

$6 x + 6 = \frac{- 36}{6}$

Divide $- 36$ by $6$ .

$6 x + 6 = - 6$

Move all terms not containing $x$ to the right side of the equation.

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Subtract $6$ from both sides of the equation.

$6 x = - 6 - 6$

Subtract $6$ from $- 6$ .

$6 x = - 12$

Divide each term by $6$ and simplify.

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Divide each term in $6 x = - 12$ by $6$ .

$\frac{6 x}{6} = \frac{- 12}{6}$

Cancel the common factor of $6$ .

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Cancel the common factor.

$\frac{6 x}{6} = \frac{- 12}{6}$

Divide $x$ by $1$ .

$x = \frac{- 12}{6}$

Divide $- 12$ by $6$ .

$x = - 2$

There is a tangent line found at $x = - 2$ parallel to the line that passes through the end points $a = - 5$ and $b = 1$ .

There is a tangent line at $x = - 2$ parallel to the line that passes through the end points $a = - 5$ and $b = 1$

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