Calculus Examples

$y = \sec (x)$

Step 1

Set $y$ as a function of $x$ .

$f (x) = \sec (x)$

Step 2

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

$\sec (x) \tan (x)$

Step 3

Set the derivative equal to $0$ then solve the equation $\sec (x) \tan (x) = 0$ .

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Step 3.1

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

$\sec (x) = 0$

$\tan (x) = 0$

Step 3.2

Set $\sec (x)$ equal to $0$ and solve for $x$ .

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Step 3.2.1

Set $\sec (x)$ equal to $0$ .

$\sec (x) = 0$

Step 3.2.2

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 3.3

Set $\tan (x)$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $\tan (x)$ equal to $0$ .

$\tan (x) = 0$

Step 3.3.2

Solve $\tan (x) = 0$ for $x$ .

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Step 3.3.2.1

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (0)$

Step 3.3.2.2

Simplify the right side.

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Step 3.3.2.2.1

The exact value of $\arctan (0)$ is $0$ .

$x = 0$

Step 3.3.2.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + 0$

Step 3.3.2.4

Add $π$ and $0$ .

$x = π$

Step 3.3.2.5

Find the period of $\tan (x)$ .

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Step 3.3.2.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 3.3.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 3.3.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 3.3.2.5.4

Divide $π$ by $1$ .

$π$

Step 3.3.2.6

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = π n, π + π n$ , for any integer $n$

Step 3.4

The final solution is all the values that make $\sec (x) \tan (x) = 0$ true.

$x = π n, π + π n$ , for any integer $n$

Step 3.5

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 4

Solve the original function $f (x) = \sec (x)$ at $x = π$ .

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Step 4.1

$f (π) = \sec (π)$

Step 4.2

Simplify the result.

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Step 4.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$f (π) = - \sec (0)$

Step 4.2.2

The exact value of $\sec (0)$ is $1$ .

$f (π) = - 1 \cdot 1$

Step 4.2.3

Multiply $- 1$ by $1$ .

$f (π) = - 1$

Step 4.2.4

The final answer is $- 1$ .

$- 1$

Step 5

The horizontal tangent line on function $f (x) = \sec (x)$ is $y = x = π n, y = - 1$ .

$y = x = π n, y = - 1$

Step 6