Calculus Examples

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

Write $\frac{x - 1}{x^{2}}$ as a function.

$f (x) = \frac{x - 1}{x^{2}}$

Find the second derivative.

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Find the first derivative.

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Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x - 1$ and $g (x) = x^{2}$ .

$f' (x) = \frac{x^{2} \frac{d}{d x} (x - 1) - (x - 1) \frac{d}{d x} x^{2}}{{(x^{2})}^{2}}$

Differentiate.

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Multiply the exponents in ${(x^{2})}^{2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $2$ by $2$ .

$f' (x) = \frac{x^{2} \frac{d}{d x} (x - 1) - (x - 1) \frac{d}{d x} x^{2}}{x^{4}}$

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$f' (x) = \frac{x^{2} (\frac{d}{d x} (x) + \frac{d}{d x} (- 1)) - (x - 1) \frac{d}{d x} x^{2}}{x^{4}}$

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

$f' (x) = \frac{x^{2} (1 + \frac{d}{d x} (- 1)) - (x - 1) \frac{d}{d x} x^{2}}{x^{4}}$

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f' (x) = \frac{x^{2} (1 + 0) - (x - 1) \frac{d}{d x} x^{2}}{x^{4}}$

Simplify the expression.

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Add $1$ and $0$ .

$f' (x) = \frac{x^{2} \cdot 1 - (x - 1) \frac{d}{d x} x^{2}}{x^{4}}$

Multiply $x^{2}$ by $1$ .

$f' (x) = \frac{x^{2} - (x - 1) \frac{d}{d x} x^{2}}{x^{4}}$

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

$f' (x) = \frac{x^{2} - (x - 1) (2 x)}{x^{4}}$

Simplify with factoring out.

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Multiply $2$ by $- 1$ .

$f' (x) = \frac{x^{2} - 2 (x - 1) x}{x^{4}}$

Factor $x$ out of $x^{2} - 2 (x - 1) x$ .

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Factor $x$ out of $x^{2}$ .

$f' (x) = \frac{x \cdot x - 2 (x - 1) x}{x^{4}}$

Factor $x$ out of $- 2 (x - 1) x$ .

$f' (x) = \frac{x \cdot x + x (- 2 (x - 1))}{x^{4}}$

Factor $x$ out of $x \cdot x + x (- 2 (x - 1))$ .

$f' (x) = \frac{x (x - 2 (x - 1))}{x^{4}}$

Cancel the common factors.

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Factor $x$ out of $x^{4}$ .

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

Cancel the common factor.

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

Rewrite the expression.

$f' (x) = \frac{x - 2 (x - 1)}{x^{3}}$

Simplify.

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Apply the distributive property.

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

Simplify the numerator.

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Multiply $- 2$ by $- 1$ .

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

Subtract $2 x$ from $x$ .

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

Factor $- 1$ out of $- x$ .

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

Rewrite $2$ as $- 1 (- 2)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 2)$ .

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

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

$f' (x) = \frac{- 1 (x - 2)}{x^{3}}$

Move the negative in front of the fraction.

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

Find the second derivative.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{x - 2}{x^{3}}$ .

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

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

Differentiate.

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Multiply the exponents in ${(x^{3})}^{2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $3$ by $2$ .

$f'' (x) = - \frac{x^{3} \frac{d}{d x} (x - 2) - (x - 2) \frac{d}{d x} x^{3}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

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

$f'' (x) = - \frac{x^{3} (\frac{d}{d x} (x) + \frac{d}{d x} (- 2)) - (x - 2) \frac{d}{d x} x^{3}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

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

$f'' (x) = - \frac{x^{3} (1 + \frac{d}{d x} (- 2)) - (x - 2) \frac{d}{d x} x^{3}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{x^{3} (1 + 0) - (x - 2) \frac{d}{d x} x^{3}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Simplify the expression.

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Add $1$ and $0$ .

$f'' (x) = - \frac{x^{3} \cdot 1 - (x - 2) \frac{d}{d x} x^{3}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Multiply $x^{3}$ by $1$ .

$f'' (x) = - \frac{x^{3} - (x - 2) \frac{d}{d x} x^{3}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

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

$f'' (x) = - \frac{x^{3} - (x - 2) (3 x^{2})}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Simplify with factoring out.

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Multiply $3$ by $- 1$ .

$f'' (x) = - \frac{x^{3} - 3 (x - 2) x^{2}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Factor $x^{2}$ out of $x^{3} - 3 (x - 2) x^{2}$ .

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Factor $x^{2}$ out of $x^{3}$ .

$f'' (x) = - \frac{x^{2} x - 3 (x - 2) x^{2}}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Factor $x^{2}$ out of $- 3 (x - 2) x^{2}$ .

$f'' (x) = - \frac{x^{2} x + x^{2} (- 3 (x - 2))}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Factor $x^{2}$ out of $x^{2} x + x^{2} (- 3 (x - 2))$ .

$f'' (x) = - \frac{x^{2} (x - 3 (x - 2))}{x^{6}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Cancel the common factors.

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Factor $x^{2}$ out of $x^{6}$ .

$f'' (x) = - \frac{x^{2} (x - 3 (x - 2))}{x^{2} x^{4}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Cancel the common factor.

$f'' (x) = - \frac{x^{2} (x - 3 (x - 2))}{x^{2} x^{4}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Rewrite the expression.

$f'' (x) = - \frac{x - 3 (x - 2)}{x^{4}} + \frac{x - 2}{x^{3}} \cdot \frac{d}{d x} (- 1)$

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{x - 3 (x - 2)}{x^{4}} + \frac{x - 2}{x^{3}} \cdot 0$

Simplify the expression.

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Multiply $\frac{x - 2}{x^{3}}$ by $0$ .

$f'' (x) = - \frac{x - 3 (x - 2)}{x^{4}} + 0$

Add $- \frac{x - 3 (x - 2)}{x^{4}}$ and $0$ .

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

Simplify.

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Apply the distributive property.

$f'' (x) = - \frac{x - 3 x - 3 \cdot - 2}{x^{4}}$

Simplify the numerator.

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Multiply $- 3$ by $- 2$ .

$f'' (x) = - \frac{x - 3 x + 6}{x^{4}}$

Subtract $3 x$ from $x$ .

$f'' (x) = - \frac{- 2 x + 6}{x^{4}}$

Factor $2$ out of $- 2 x + 6$ .

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Factor $2$ out of $- 2 x$ .

$f'' (x) = - \frac{2 (- x) + 6}{x^{4}}$

Factor $2$ out of $6$ .

$f'' (x) = - \frac{2 (- x) + 2 (3)}{x^{4}}$

Factor $2$ out of $2 (- x) + 2 (3)$ .

$f'' (x) = - \frac{2 (- x + 3)}{x^{4}}$

Factor $- 1$ out of $- x$ .

$f'' (x) = - \frac{2 (- (x) + 3)}{x^{4}}$

Rewrite $3$ as $- 1 (- 3)$ .

$f'' (x) = - \frac{2 (- (x) - 1 \cdot - 3)}{x^{4}}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

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

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

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

Move the negative in front of the fraction.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{2 (x - 3)}{x^{4}}$ by $1$ .

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

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

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

Set the second derivative equal to $0$ then solve the equation $\frac{2 (x - 3)}{x^{4}} = 0$ .

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Set the second derivative equal to $0$ .

$\frac{2 (x - 3)}{x^{4}} = 0$

Set the numerator equal to zero.

$2 (x - 3) = 0$

Solve the equation for $x$ .

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Divide each term in $2 (x - 3) = 0$ by $2$ and simplify.

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Divide each term in $2 (x - 3) = 0$ by $2$ .

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

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x - 3$ by $1$ .

$x - 3 = \frac{0}{2}$

Simplify the right side.

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Divide $0$ by $2$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Find the points where the second derivative is $0$ .

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Substitute $3$ in $f (x) = \frac{x - 1}{x^{2}}$ to find the value of $y$ .

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Replace the variable $x$ with $3$ in the expression.

$f (3) = \frac{(3) - 1}{{(3)}^{2}}$

Simplify the result.

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Subtract $1$ from $3$ .

$f (3) = \frac{2}{3^{2}}$

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

$f (3) = \frac{2}{9}$

The final answer is $\frac{2}{9}$ .

$\frac{2}{9}$

The point found by substituting $3$ in $f (x) = \frac{x - 1}{x^{2}}$ is $(3, \frac{2}{9})$ . This point can be an inflection point.

$(3, \frac{2}{9})$

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, 3) \cup (3, \infty)$

Substitute a value from the interval $(- \infty, 3)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $2.9$ in the expression.

$f'' (2.9) = \frac{2 ((2.9) - 3)}{{(2.9)}^{4}}$

Simplify the result.

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Subtract $3$ from $2.9$ .

$f'' (2.9) = \frac{2 \cdot - 0.1}{{2.9}^{4}}$

Raise $2.9$ to the power of $4$ .

$f'' (2.9) = \frac{2 \cdot - 0.1}{70.7281}$

Multiply $2$ by $- 0.1$ .

$f'' (2.9) = \frac{- 0.2}{70.7281}$

Divide $- 0.2$ by $70.7281$ .

$f'' (2.9) = - 0.00282773$

The final answer is $- 0.00282773$ .

$- 0.00282773$

At $2.9$ , the second derivative is $- 0.00282773$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 3)$

Decreasing on $(- \infty, 3)$ since $f'' (x) < 0$

Substitute a value from the interval $(3, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $3.1$ in the expression.

$f'' (3.1) = \frac{2 ((3.1) - 3)}{{(3.1)}^{4}}$

Simplify the result.

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Subtract $3$ from $3.1$ .

$f'' (3.1) = \frac{2 \cdot 0.1}{{3.1}^{4}}$

Raise $3.1$ to the power of $4$ .

$f'' (3.1) = \frac{2 \cdot 0.1}{92.3521}$

Multiply $2$ by $0.1$ .

$f'' (3.1) = \frac{0.2}{92.3521}$

Divide $0.2$ by $92.3521$ .

$f'' (3.1) = 0.00216562$

The final answer is $0.00216562$ .

$0.00216562$

At $3.1$ , the second derivative is $0.00216562$ . Since this is positive, the second derivative is increasing on the interval $(3, \infty)$ .

Increasing on $(3, \infty)$ since $f'' (x) > 0$

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(3, \frac{2}{9})$ .

$(3, \frac{2}{9})$