Calculus Examples

$f (x) = x^{2} + 2 x - 3$ , $[0, 6]$

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 $f (x) = x^{2} + 2 x - 3$ 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 $[0, 6]$ .

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

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

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

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

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 $2 \frac{d}{d x} [x]$ .

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

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

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

Multiply $2$ by $1$ to get $2$ .

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

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

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

Add $2 x + 2$ and $0$ to get $2 x + 2$ .

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

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

$2 x + 2$

Find if the derivative is continuous on $(0, 6)$ .

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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.

$(- \infty, \infty)$

${x | x \in ℝ}$

$f' (x)$ is continuous on $(0, 6)$ .

The function is continuous.

The function is differentiable on $(0, 6)$ because the derivative is continuous on $(0, 6)$ .

The function is differentiable.

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[0, 6]$ and differentiable on $(0, 6)$ .

$f (x)$ is continuous on $[0, 6]$ and differentiable on $(0, 6)$ .

Evaluate $f (a)$ from the interval $[0, 6]$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{2} + 2 (0) - 3$

Simplify the result.

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Simplify each term.

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Remove parentheses around $0$ .

$f (0) = 0^{2} + 2 (0) - 3$

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 2 (0) - 3$

Multiply $2$ by $0$ to get $0$ .

$f (0) = 0 + 0 - 3$

Simplify by adding zeros.

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Add $0$ and $0$ to get $0$ .

$f (0) = 0 - 3$

Subtract $3$ from $0$ to get $- 3$ .

$f (0) = - 3$

The final answer is $- 3$ .

$- 3$

Evaluate $f (b)$ from the interval $[0, 6]$ .

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Replace the variable $x$ with $6$ in the expression.

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

Simplify the result.

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Simplify each term.

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Remove parentheses around $6$ .

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

Raise $6$ to the power of $2$ to get $36$ .

$f (6) = 36 + 2 (6) - 3$

Multiply $2$ by $6$ to get $12$ .

$f (6) = 36 + 12 - 3$

Simplify by adding and subtracting.

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Add $36$ and $12$ to get $48$ .

$f (6) = 48 - 3$

Subtract $3$ from $48$ to get $45$ .

$f (6) = 45$

The final answer is $45$ .

$45$

Solve $2 x + 2 = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $2 x + 2 = \frac{- (f (6) + f (0))}{- (6 + 0)}$ .

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Simplify the right side.

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Reduce the expression by cancelling the common factors.

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Factor $3$ out of $- 1 (- 6 + 0)$ .

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Factor $3$ out of $- 6$ .

$2 x + 2 = \frac{3 (15 + 1)}{- 1 (3 \cdot - 2 + 0)}$

Factor $3$ out of $0$ .

$2 x + 2 = \frac{3 (15 + 1)}{- 1 (3 \cdot - 2 + 3 \cdot 0)}$

Factor $3$ out of $3 \cdot - 2 + 3 \cdot 0$ .

$2 x + 2 = \frac{3 (15 + 1)}{- 1 (3 \cdot (- 2 + 0))}$

Move $3$ .

$2 x + 2 = \frac{3 (15 + 1)}{3 (- 1 (- 2 + 0))}$

Cancel the common factor.

$2 x + 2 = \frac{3 (15 + 1)}{3 (- 1 (- 2 + 0))}$

Rewrite the expression.

$2 x + 2 = \frac{15 + 1}{- 1 (- 2 + 0)}$

Simplify the numerator.

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Multiply $- 1$ by $- 1$ to get $1$ .

$2 x + 2 = \frac{15 + 1}{- 1 (- 2 + 0)}$

Add $15$ and $1$ to get $16$ .

$2 x + 2 = \frac{16}{- 1 (- 2 + 0)}$

Simplify the denominator.

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Factor $3$ out of $- 6$ .

$2 x + 2 = \frac{16}{- 1 (3 (- 2) + 0)}$

Factor $3$ out of $0$ .

$2 x + 2 = \frac{16}{- 1 (3 \cdot - 2 + 3 \cdot 0)}$

Factor $3$ out of $3 \cdot - 2 + 3 \cdot 0$ .

$2 x + 2 = \frac{16}{- 1 (3 \cdot (- 2 + 0))}$

Rewrite.

$2 x + 2 = \frac{16}{- 1 (- 2 + 0)}$

Add $- 2$ and $0$ to get $- 2$ .

$2 x + 2 = \frac{16}{- 1 \cdot - 2}$

Reduce the expression by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $2$ out of $- 1 \cdot - 2$ .

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Factor $2$ out of $- 2$ .

$2 x + 2 = \frac{2 \cdot 8}{- 1 \cdot (2 \cdot - 1)}$

Factor $2$ out of $- 2$ .

$2 x + 2 = \frac{2 (8)}{- 1 \cdot (2 (- 1))}$

Move $2$ .

$2 x + 2 = \frac{2 (8)}{2 (- 1 \cdot - 1)}$

Multiply $- 1$ by $- 1$ to get $- 1 \cdot - 1$ .

$2 x + 2 = \frac{2 (8)}{2 (- 1 \cdot - 1)}$

Cancel the common factor.

$2 x + 2 = \frac{2 \cdot 8}{2 (- 1 \cdot - 1)}$

Rewrite the expression.

$2 x + 2 = \frac{8}{- 1 \cdot - 1}$

Simplify the expression.

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Multiply $- 1$ by $- 1$ to get $1$ .

$2 x + 2 = \frac{8}{1}$

Divide $8$ by $1$ to get $8$ .

$2 x + 2 = 8$

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

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Subtract $2$ from both sides of the equation.

$2 x = - 2 + 8$

Add $- 2$ and $8$ to get $6$ .

$2 x = 6$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 6$ by $2$ .

$\frac{2 x}{2} = \frac{6}{2}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{2 x}{2} = \frac{6}{2}$

Divide $x$ by $1$ to get $x$ .

$x = \frac{6}{2}$

Divide $6$ by $2$ to get $3$ .

$x = 3$

There is a tangent line found at $x = 3$ parallel to the line that passes through the end points $a = 0$ and $b = 6$ .

There is a tangent line at $x = 3$ parallel to the line that passes through the end points $a = 0$ and $b = 6$

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