Calculus Examples

$f (x) = x^{4} - 3 x^{3} + 4$ , $[1, 2]$

Step 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 $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

Check if $f (x) = x^{4} - 3 x^{3} + 4$ is continuous.

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Step 2.1

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:

${x | x \in ℝ}$

Step 2.2

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

The function is continuous.

Step 3

Find the derivative.

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Step 3.1

Find the first derivative.

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Step 3.1.1

Differentiate.

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Step 3.1.1.1

By the Sum Rule, the derivative of $x^{4} - 3 x^{3} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [4]$

Step 3.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [4]$

Step 3.1.2

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

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Step 3.1.2.1

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

$4 x^{3} - 3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [4]$

Step 3.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$4 x^{3} - 3 (3 x^{2}) + \frac{d}{d x} [4]$

Step 3.1.2.3

Multiply $3$ by $- 3$ .

$4 x^{3} - 9 x^{2} + \frac{d}{d x} [4]$

Step 3.1.3

Differentiate using the Constant Rule.

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Step 3.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$4 x^{3} - 9 x^{2} + 0$

Step 3.1.3.2

Add $4 x^{3} - 9 x^{2}$ and $0$ .

$f' (x) = 4 x^{3} - 9 x^{2}$

Step 3.2

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 9 x^{2}$ .

$4 x^{3} - 9 x^{2}$

Step 4

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

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Step 4.1

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:

${x | x \in ℝ}$

Step 4.2

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

The function is continuous.

Step 5

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

The function is differentiable.

Step 6

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

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

Step 7

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

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Step 7.1

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

$f (1) = {(1)}^{4} - 3 {(1)}^{3} + 4$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

One to any power is one.

$f (1) = 1 - 3 {(1)}^{3} + 4$

Step 7.2.1.2

One to any power is one.

$f (1) = 1 - 3 \cdot 1 + 4$

Step 7.2.1.3

Multiply $- 3$ by $1$ .

$f (1) = 1 - 3 + 4$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Subtract $3$ from $1$ .

$f (1) = - 2 + 4$

Step 7.2.2.2

Add $- 2$ and $4$ .

$f (1) = 2$

Step 7.2.3

The final answer is $2$ .

$2$

Step 8

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

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Step 8.1

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

$f (2) = {(2)}^{4} - 3 {(2)}^{3} + 4$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $2$ to the power of $4$ .

$f (2) = 16 - 3 {(2)}^{3} + 4$

Step 8.2.1.2

Raise $2$ to the power of $3$ .

$f (2) = 16 - 3 \cdot 8 + 4$

Step 8.2.1.3

Multiply $- 3$ by $8$ .

$f (2) = 16 - 24 + 4$

Step 8.2.2

Simplify by adding and subtracting.

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Step 8.2.2.1

Subtract $24$ from $16$ .

$f (2) = - 8 + 4$

Step 8.2.2.2

Add $- 8$ and $4$ .

$f (2) = - 4$

Step 8.2.3

The final answer is $- 4$ .

$- 4$

Step 9

Solve $4 x^{3} - 9 x^{2} = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $4 x^{3} - 9 x^{2} = \frac{- (f (2) + f (1))}{- (2 + 1)}$ .

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Step 9.1

Simplify $\frac{(- 4) - (2)}{(2) - (1)}$ .

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Step 9.1.1

Simplify the numerator.

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Step 9.1.1.1

Multiply $- 1$ by $2$ .

$4 x^{3} - 9 x^{2} = \frac{- 4 - 2}{2 - (1)}$

Step 9.1.1.2

Subtract $2$ from $- 4$ .

$4 x^{3} - 9 x^{2} = \frac{- 6}{2 - (1)}$

Step 9.1.2

Simplify the denominator.

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Step 9.1.2.1

Multiply $- 1$ by $1$ .

$4 x^{3} - 9 x^{2} = \frac{- 6}{2 - 1}$

Step 9.1.2.2

Subtract $1$ from $2$ .

$4 x^{3} - 9 x^{2} = \frac{- 6}{1}$

Step 9.1.3

Divide $- 6$ by $1$ .

$4 x^{3} - 9 x^{2} = - 6$

Step 9.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 0.711668, 1.18903787, 1.77263012$

Step 10

There is a tangent line found at $x = - 0.711668$ parallel to the line that passes through the end points $a = 1$ and $b = 2$ .

There is a tangent line at $x = - 0.711668$ parallel to the line that passes through the end points $a = 1$ and $b = 2$

Step 11

There is a tangent line found at $x = 1.18903787$ parallel to the line that passes through the end points $a = 1$ and $b = 2$ .

There is a tangent line at $x = 1.18903787$ parallel to the line that passes through the end points $a = 1$ and $b = 2$

Step 12

There is a tangent line found at $x = 1.77263012$ parallel to the line that passes through the end points $a = 1$ and $b = 2$ .

There is a tangent line at $x = 1.77263012$ parallel to the line that passes through the end points $a = 1$ and $b = 2$

Step 13