Calculus Examples

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

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.

$(- \infty, \infty)$

${x | x \in ℝ}$

$f (x)$ is continuous on $[- 2, 1]$ .

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 $1$ .

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

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 $6 x + 6$ .

$6 x + 6$

Find if the derivative is continuous on $(- 2, 1)$ .

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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.

$(- \infty, \infty)$

${x | x \in ℝ}$

$f' (x)$ is continuous on $(- 2, 1)$ .

The function is continuous.

The function is differentiable on $(- 2, 1)$ because the derivative is continuous on $(- 2, 1)$ .

The function is differentiable.

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[- 2, 1]$ and differentiable on $(- 2, 1)$ .

$f (x)$ is continuous on $[- 2, 1]$ and differentiable on $(- 2, 1)$ .

Evaluate $f (a)$ from the interval $[- 2, 1]$ .

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Replace the variable $x$ with $- 2$ in the expression.

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

Simplify the result.

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Simplify each term.

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

$f (- 2) = 3 \cdot 4 + 6 (- 2) - 5$

Multiply $3$ by $4$ .

$f (- 2) = 12 + 6 (- 2) - 5$

Multiply $6$ by $- 2$ .

$f (- 2) = 12 - 12 - 5$

Simplify by subtracting numbers.

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Subtract $12$ from $12$ .

$f (- 2) = 0 - 5$

Subtract $5$ from $0$ .

$f (- 2) = - 5$

The final answer is $- 5$ .

$- 5$

Evaluate $f (b)$ from the interval $[- 2, 1]$ .

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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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Remove parentheses around $1$ .

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

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 $x$ . $6 x + 6 = \frac{- (f (1) + f (- 2))}{- (1 - 2)}$ .

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Simplify the right side.

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Simplify the numerator.

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Multiply $- 1$ by $- 5$ .

$6 x + 6 = \frac{4 + 5}{1 - (- 2)}$

Add $4$ and $5$ .

$6 x + 6 = \frac{9}{1 - (- 2)}$

Simplify the denominator.

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Multiply $- 1$ by $- 2$ .

$6 x + 6 = \frac{9}{1 + 2}$

Add $1$ and $2$ .

$6 x + 6 = \frac{9}{3}$

Divide $9$ by $3$ .

$6 x + 6 = 3$

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.

$6 x = - 6 + 3$

Add $- 6$ and $3$ .

$6 x = - 3$

Divide each term by $6$ and simplify.

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Divide each term in $6 x = - 3$ by $6$ .

$\frac{6 x}{6} = - \frac{3}{6}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{6 x}{6} = - \frac{3}{6}$

Divide $x$ by $1$ .

$x = - \frac{3}{6}$

Reduce the expression by cancelling the common factors.

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Factor $3$ out of $3$ .

$x = - \frac{3 (1)}{6}$

Cancel the common factors.

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Factor $3$ out of $6$ .

$x = - \frac{3 \cdot 1}{3 \cdot 2}$

Cancel the common factor.

$x = - \frac{3 \cdot 1}{3 \cdot 2}$

Rewrite the expression.

$x = - \frac{1}{2}$

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

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

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