Calculus Examples

$e^{2 x} - 3 e^{x} + 2 < 0$

Step 1

Rewrite $e^{2 x}$ as exponentiation.

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

Step 2

Substitute $u$ for $e^{x}$ .

$u^{2} - 3 u + 2 = 0$

Step 3

Solve for $u$ .

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Step 3.1

Factor $u^{2} - 3 u + 2$ using the AC method.

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Step 3.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $- 3$ .

$- 2, - 1$

Step 3.1.2

Write the factored form using these integers.

$(u - 2) (u - 1) = 0$

Step 3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u - 2 = 0$

$u - 1 = 0$

Step 3.3

Set $u - 2$ equal to $0$ and solve for $u$ .

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Step 3.3.1

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Step 3.3.2

Add $2$ to both sides of the equation.

$u = 2$

Step 3.4

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 3.4.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 3.4.2

Add $1$ to both sides of the equation.

$u = 1$

Step 3.5

The final solution is all the values that make $(u - 2) (u - 1) = 0$ true.

$u = 2, 1$

Step 4

Substitute $2$ for $u$ in $u = e^{x}$ .

$2 = e^{x}$

Step 5

Solve $2 = e^{x}$ .

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Step 5.1

Rewrite the equation as $e^{x} = 2$ .

$e^{x} = 2$

Step 5.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 5.3

Expand the left side.

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Step 5.3.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (2)$

Step 5.3.2

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

$x \cdot 1 = \ln (2)$

Step 5.3.3

Multiply $x$ by $1$ .

$x = \ln (2)$

Step 6

Substitute $1$ for $u$ in $u = e^{x}$ .

$1 = e^{x}$

Step 7

Solve $1 = e^{x}$ .

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Step 7.1

Rewrite the equation as $e^{x} = 1$ .

$e^{x} = 1$

Step 7.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 7.3

Expand the left side.

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Step 7.3.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (1)$

Step 7.3.2

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

$x \cdot 1 = \ln (1)$

Step 7.3.3

Multiply $x$ by $1$ .

$x = \ln (1)$

Step 7.4

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

$x = 0$

Step 8

List the solutions that makes the equation true.

$x = \ln (2), 0$

Step 9

Use each root to create test intervals.

$x < 0$

$0 < x < \ln (2)$

$x > \ln (2)$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 10.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 10.1.2

Replace $x$ with $- 2$ in the original inequality.

$e^{2 (- 2)} - 3 e^{- 2} + 2 < 0$

Step 10.1.3

The left side $1.61230978$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $0 < x < \ln (2)$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $0 < x < \ln (2)$ and see if this value makes the original inequality true.

$x = 0.35$

Step 10.2.2

Replace $x$ with $0.35$ in the original inequality.

$e^{2 (0.35)} - 3 e^{0.35} + 2 < 0$

Step 10.2.3

The left side $- 0.24344993$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $x > \ln (2)$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > \ln (2)$ and see if this value makes the original inequality true.

$x = 3$

Step 10.3.2

Replace $x$ with $3$ in the original inequality.

$e^{2 (3)} - 3 e^{3} + 2 < 0$

Step 10.3.3

The left side $345.17218272$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 10.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < \ln (2)$ True

$x > \ln (2)$ False

$x < 0$ False

$0 < x < \ln (2)$ True

$x > \ln (2)$ False

Step 11

The solution consists of all of the true intervals.

$0 < x < \ln (2)$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$0 < x < \ln (2)$

Interval Notation:

$(0, \ln (2))$

Step 13