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.

Tap for more steps...

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 $[0, 6]$ .

The function is continuous.

Find the derivative.

Tap for more steps...

Find the first derivative.

Tap for more steps...

Differentiate.

Tap for more steps...

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

Tap for more steps...

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 $1$ .

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

Differentiate using the Constant Rule.

Tap for more steps...

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 $0$ .

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

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

$2 x + 2$

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

Tap for more steps...

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 $(0, 6)$ .

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 $[0, 6]$ .

Tap for more steps...

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

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

Simplify the result.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

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

Multiply $2$ by $0$ .

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

Simplify by adding zeros.

Tap for more steps...

Add $0$ and $0$ .

$f (0) = 0 - 3$

Subtract $3$ from $0$ .

$f (0) = - 3$

The final answer is $- 3$ .

$- 3$

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

Tap for more steps...

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

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

Simplify the result.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

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

Multiply $2$ by $6$ .

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

Simplify by adding and subtracting.

Tap for more steps...

Add $36$ and $12$ .

$f (6) = 48 - 3$

Subtract $3$ from $48$ .

$f (6) = 45$

The final answer is $45$ .

$45$

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

Tap for more steps...

Simplify $\frac{45 - (- 3)}{6 - (0)}$ .

Tap for more steps...

Reduce the expression by cancelling the common factors.

Tap for more steps...

Rewrite $6$ as $- 1 (- 6)$ .

$2 x + 2 = \frac{45 - (- 3)}{- 1 \cdot - 6 - (0)}$

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

$2 x + 2 = \frac{45 - (- 3)}{- 1 (- 6 + 0)}$

Reorder terms.

$2 x + 2 = \frac{45 - 3 \cdot - 1}{- 1 (- 6 + 0)}$

Factor $3$ out of $45$ .

$2 x + 2 = \frac{3 (15) - 3 \cdot - 1}{- 1 (- 6 + 0)}$

Factor $3$ out of $- 3 \cdot - 1$ .

$2 x + 2 = \frac{3 (15) + 3 (1)}{- 1 (- 6 + 0)}$

Factor $3$ out of $3 (15) + 3 (- - 1)$ .

$2 x + 2 = \frac{3 (15 + 1)}{- 1 (- 6 + 0)}$

Cancel the common factors.

Tap for more steps...

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

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

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.

Tap for more steps...

Multiply $- 1$ by $- 1$ .

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

Add $15$ and $1$ .

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

Simplify the expression.

Tap for more steps...

Add $- 2$ and $0$ .

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

Multiply $- 1$ by $- 2$ .

$2 x + 2 = \frac{16}{2}$

Divide $16$ by $2$ .

$2 x + 2 = 8$

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

Tap for more steps...

Subtract $2$ from both sides of the equation.

$2 x = 8 - 2$

Subtract $2$ from $8$ .

$2 x = 6$

Divide each term by $2$ and simplify.

Tap for more steps...

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

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

Reduce the expression by cancelling the common factors.

Tap for more steps...

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Divide $6$ by $2$ .

$x = 3$

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 $b = 6$

Enter YOUR Problem