Question

Use the guidelines of section 4.4 to sketch the curve

22.\(y = \sqrt {1 - x} + \sqrt {1 + x} \)

Short answer

The graph of the given function is drawn.

Step-by-step solution

Domain of \(y\)

Consider the given function as \(f\left( x \right) = \sqrt {1 - x} + \sqrt {1 + x} \).

Here, \(1 - x \ge 0\) or \(1 + x \ge 0\) implies \(x \le 1\) or \(x \ge 1\).

Thus, the domain of y is \( - 1 \le x \le 1\).

Intercepts of \(y\)

The y-intercept is the value of y at x=0. Similarly, x-intercept is the value of x at y=0.

Let x=0 in y then,\(y\left( 0 \right) = 2\)implies y-intercept is 2. Similarly, let y=0 in \(0 = \sqrt {1 - x} + \sqrt {1 + x} \)implies there is no x intercept.

Therefore, given function has no x intercept and y intercept at \(y = 2\).

Symmetry of \(y\)

Consider the given function as \(f\left( x \right) = \sqrt {1 - x} + \sqrt {1 + x} \).

Substitute x by –x then,

\(\begin{array}{c}f\left( { - x} \right) = \sqrt {1 - \left( { - x} \right)} + \sqrt {1 - x} \\ = \sqrt {1 - x} + \sqrt {1 + x} \\ = f\left( x \right)\end{array}\)

Here, \(f\left( { - x} \right) = f\left( x \right)\) implies f is an even function.

Asymptotes of \(y\)

Since, there is no denominator, there is no vertical asymptote and no horizontal asymptote.

Local maximum and minimum

The first derivative of the given function is\(f'\left( x \right) = \frac{{ - 1}}{{2\sqrt {1 - x} }} + \frac{1}{{2\sqrt {1 + x} }}\).

The derivative is zero at \(x = - 1,1\).

Thus, the function is increasing on \(\left( { - 1,0} \right)\) and decreasing on \(\left( {0,1} \right)\).

Concavity and inflection point

The second derivative of the given function is\(f'\left( x \right) = \frac{{ - 1}}{{4{{\left( {1 - x} \right)}^{\frac{3}{2}}}}} - \frac{1}{{4{{\left( {1 + x} \right)}^{\frac{3}{2}}}}}\).

The second derivative is 0 at the point\(x = - 1,1\).

The graph is concave downward on the interval \(\left( { - 1,1} \right)\).

Graph of \(y\)

Draw the graph of the given function as shown below:

Therefore, the graph of the given function is drawn.