Calculus Examples

f (x) = ln (x^{2} + 1)

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

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in

0

$0$ for

y

$y$ and solve for

x

$x$ .

0 = ln (x^{2} + 1)

$0 = \ln (x^{2} + 1)$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as

ln (x^{2} + 1) = 0

$\ln (x^{2} + 1) = 0$ .

ln (x^{2} + 1) = 0

$\ln (x^{2} + 1) = 0$

Step 1.2.2

To solve for

x

$x$ , rewrite the equation using properties of logarithms.

e^{ln (x^{2} + 1)} = e^{0}

$e^{\ln (x^{2} + 1)} = e^{0}$

Step 1.2.3

Rewrite

ln (x^{2} + 1) = 0

$\ln (x^{2} + 1) = 0$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

{log}_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

e^{0} = x^{2} + 1

$e^{0} = x^{2} + 1$

Step 1.2.4

Solve for

x

$x$ .

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Step 1.2.4.1

Rewrite the equation as

x^{2} + 1 = e^{0}

$x^{2} + 1 = e^{0}$ .

x^{2} + 1 = e^{0}

$x^{2} + 1 = e^{0}$

Step 1.2.4.2

Anything raised to

0

$0$ is

1

$1$ .

x^{2} + 1 = 1

$x^{2} + 1 = 1$

Step 1.2.4.3

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 1.2.4.3.1

Subtract

1

$1$ from both sides of the equation.

x^{2} = 1 - 1

$x^{2} = 1 - 1$

Step 1.2.4.3.2

Subtract

1

$1$ from

1

$1$ .

x^{2} = 0

$x^{2} = 0$

x^{2} = 0

$x^{2} = 0$

Step 1.2.4.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

x = \pm \sqrt{0}

$x = \pm \sqrt{0}$

Step 1.2.4.5

Simplify

\pm \sqrt{0}

$\pm \sqrt{0}$ .

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Step 1.2.4.5.1

Rewrite

0

$0$ as

0^{2}

$0^{2}$ .

x = \pm \sqrt{0^{2}}

$x = \pm \sqrt{0^{2}}$

Step 1.2.4.5.2

Pull terms out from under the radical, assuming positive real numbers.

x = \pm 0

$x = \pm 0$

Step 1.2.4.5.3

Plus or minus

0

$0$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

x = 0

$x = 0$

x = 0

$x = 0$

Step 1.3

x-intercept(s) in point form.

x-intercept(s):

(0, 0)

$(0, 0)$

x-intercept(s):

(0, 0)

$(0, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in

0

$0$ for

x

$x$ and solve for

y

$y$ .

y = ln ({(0)}^{2} + 1)

$y = \ln ({(0)}^{2} + 1)$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

y = ln (0^{2} + 1)

$y = \ln (0^{2} + 1)$

Step 2.2.2

Remove parentheses.

y = ln ({(0)}^{2} + 1)

$y = \ln ({(0)}^{2} + 1)$

Step 2.2.3

Simplify

ln ({(0)}^{2} + 1)

$\ln ({(0)}^{2} + 1)$ .

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Step 2.2.3.1

Raising

0

$0$ to any positive power yields

0

$0$ .

y = ln (0 + 1)

$y = \ln (0 + 1)$

Step 2.2.3.2

Add

0

$0$ and

1

$1$ .

y = ln (1)

$y = \ln (1)$

Step 2.2.3.3

The natural logarithm of

1

$1$ is

0

$0$ .

y = 0

$y = 0$

y = 0

$y = 0$

y = 0

$y = 0$

Step 2.3

y-intercept(s) in point form.

y-intercept(s):

(0, 0)

$(0, 0)$

y-intercept(s):

(0, 0)

$(0, 0)$

Step 3

List the intersections.

x-intercept(s):

(0, 0)

$(0, 0)$

y-intercept(s):

(0, 0)

$(0, 0)$

Step 4