Pre-Algebra Examples

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

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\frac{2 (1 - \ln (x))}{x^{2}}$ is undefined.

$x \leq 0$

Step 1.2

Since $\frac{2 (1 - \ln (x))}{x^{2}}$ $\to$ $\infty$ as $x$ $\to$ $0$ from the left and $\frac{2 (1 - \ln (x))}{x^{2}}$ $\to$ $- \infty$ as $x$ $\to$ $0$ from the right, then $x = 0$ is a vertical asymptote.

$x = 0$

Step 1.3

Evaluate $lim_{x \to \infty} \frac{2 (1 - \ln (x))}{x^{2}}$ to find the horizontal asymptote.

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Step 1.3.1

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to \infty} \frac{1 - \ln (x)}{x^{2}}$

Step 1.3.2

Apply L'Hospital's rule.

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Step 1.3.2.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.3.2.1.1

Take the limit of the numerator and the limit of the denominator.

$2 \frac{lim_{x \to \infty} 1 - \ln (x)}{lim_{x \to \infty} x^{2}}$

Step 1.3.2.1.2

Evaluate the limit of the numerator.

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Step 1.3.2.1.2.1

Evaluate the limit.

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Step 1.3.2.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\infty$ .

$2 \frac{lim_{x \to \infty} 1 - lim_{x \to \infty} \ln (x)}{lim_{x \to \infty} x^{2}}$

Step 1.3.2.1.2.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $\infty$ .

$2 \frac{1 - lim_{x \to \infty} \ln (x)}{lim_{x \to \infty} x^{2}}$

Step 1.3.2.1.2.2

As log approaches infinity, the value goes to $\infty$ .

$2 \frac{1 - \infty}{lim_{x \to \infty} x^{2}}$

Step 1.3.2.1.2.3

Simplify the answer.

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Step 1.3.2.1.2.3.1

A non-zero constant times infinity is infinity.

$2 \frac{1 + - \infty}{lim_{x \to \infty} x^{2}}$

Step 1.3.2.1.2.3.2

Infinity plus or minus a number is infinity.

$2 \frac{- \infty}{lim_{x \to \infty} x^{2}}$

Step 1.3.2.1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$2 \frac{- \infty}{\infty}$

Step 1.3.2.1.4

Infinity divided by infinity is undefined.

Undefined

$2 \frac{- \infty}{\infty}$

Step 1.3.2.2

Since $\frac{- \infty}{\infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to \infty} \frac{1 - \ln (x)}{x^{2}} = lim_{x \to \infty} \frac{\frac{d}{d x} [1 - \ln (x)]}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3

Find the derivative of the numerator and denominator.

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Step 1.3.2.3.1

Differentiate the numerator and denominator.

$2 lim_{x \to \infty} \frac{\frac{d}{d x} [1 - \ln (x)]}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3.2

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

$2 lim_{x \to \infty} \frac{\frac{d}{d x} [1] + \frac{d}{d x} [- \ln (x)]}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3.3

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

$2 lim_{x \to \infty} \frac{0 + \frac{d}{d x} [- \ln (x)]}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3.4

Evaluate $\frac{d}{d x} [- \ln (x)]$ .

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Step 1.3.2.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \ln (x)$ with respect to $x$ is $- \frac{d}{d x} [\ln (x)]$ .

$2 lim_{x \to \infty} \frac{0 - \frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3.4.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$2 lim_{x \to \infty} \frac{0 - \frac{1}{x}}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3.5

Subtract $\frac{1}{x}$ from $0$ .

$2 lim_{x \to \infty} \frac{- \frac{1}{x}}{\frac{d}{d x} [x^{2}]}$

Step 1.3.2.3.6

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

$2 lim_{x \to \infty} \frac{- \frac{1}{x}}{2 x}$

Step 1.3.2.4

Multiply the numerator by the reciprocal of the denominator.

$2 lim_{x \to \infty} - \frac{1}{x} \cdot \frac{1}{2 x}$

Step 1.3.2.5

Combine factors.

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Step 1.3.2.5.1

Multiply $\frac{1}{2 x}$ by $\frac{1}{x}$ .

$2 lim_{x \to \infty} - \frac{1}{2 x \cdot x}$

Step 1.3.2.5.2

Raise $x$ to the power of $1$ .

$2 lim_{x \to \infty} - \frac{1}{2 x^{1} x}$

Step 1.3.2.5.3

Raise $x$ to the power of $1$ .

$2 lim_{x \to \infty} - \frac{1}{2 x^{1} x^{1}}$

Step 1.3.2.5.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 lim_{x \to \infty} - \frac{1}{2 x^{1 + 1}}$

Step 1.3.2.5.5

Add $1$ and $1$ .

$2 lim_{x \to \infty} - \frac{1}{2 x^{2}}$

Step 1.3.3

Evaluate the limit.

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Step 1.3.3.1

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

$2 \cdot - 1 lim_{x \to \infty} \frac{1}{2 x^{2}}$

Step 1.3.3.2

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$2 \cdot - 1 \frac{1}{2} lim_{x \to \infty} \frac{1}{x^{2}}$

Step 1.3.4

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

$2 \cdot - 1 \frac{1}{2} \cdot 0$

Step 1.3.5

Simplify the answer.

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Step 1.3.5.1

Cancel the common factor of $2$ .

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Step 1.3.5.1.1

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

$2 (- 1) \frac{1}{2} \cdot 0$

Step 1.3.5.1.2

Cancel the common factor.

$2 \cdot - 1 \frac{1}{2} \cdot 0$

Step 1.3.5.1.3

Rewrite the expression.

$- 1 \cdot 0$

Step 1.3.5.2

Multiply $- 1$ by $0$ .

$0$

Step 1.4

List the horizontal asymptotes:

$y = 0$

Step 1.5

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.6

This is the set of all asymptotes.

Vertical Asymptotes: $x = 0$

Horizontal Asymptotes: $y = 0$

Vertical Asymptotes: $x = 0$

Horizontal Asymptotes: $y = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify the numerator.

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Step 2.2.1.1

The natural logarithm of $1$ is $0$ .

$f (1) = \frac{2 (1 - 0)}{1^{2}}$

Step 2.2.1.2

Multiply $- 1$ by $0$ .

$f (1) = \frac{2 (1 + 0)}{1^{2}}$

Step 2.2.1.3

Add $1$ and $0$ .

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

Step 2.2.2

Simplify the expression.

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Step 2.2.2.1

One to any power is one.

$f (1) = \frac{2 \cdot 1}{1}$

Step 2.2.2.2

Multiply $2$ by $1$ .

$f (1) = \frac{2}{1}$

Step 2.2.2.3

Divide $2$ by $1$ .

$f (1) = 2$

Step 2.2.3

The final answer is $2$ .

$2$

Step 2.3

Convert $2$ to decimal.

$y = 2$

Step 3

Find the point at $x = 2$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Cancel the common factors.

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Step 3.2.1.1

Factor $2$ out of ${(2)}^{2}$ .

$f (2) = \frac{(2) (1 - \ln (2))}{2 \cdot 2}$

Step 3.2.1.2

Cancel the common factor.

$f (2) = \frac{2 (1 - \ln (2))}{2 \cdot 2}$

Step 3.2.1.3

Rewrite the expression.

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

Step 3.2.2

The final answer is $\frac{1 - \ln (2)}{2}$ .

$\frac{1 - \ln (2)}{2}$

Step 3.3

Convert $\frac{1 - \ln (2)}{2}$ to decimal.

$y = 0.1534264$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

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

$f (3) = \frac{2 (1 - \ln (3))}{9}$

Step 4.2.2

The final answer is $\frac{2 (1 - \ln (3))}{9}$ .

$\frac{2 (1 - \ln (3))}{9}$

Step 4.3

Convert $\frac{2 (1 - \ln (3))}{9}$ to decimal.

$y = - 0.02191384$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 2), (2, 0.1534264), (3, - 0.02191384)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 2 \\ 2 & 0.153 \\ 3 & - 0.022 \end{array}$

Step 6