Question

1 –54 Use the guidelines of this section to sketch the curve.

34. \(y = x + \cos x\)

Short answer

Graph of the given curve are:

Step-by-step solution

Steps for sketching a curve

There are following terms needed to examine forSketching a Graphfor any given Function:

1. Find the Domainof the Function.
2. Calculate the Intercepts.
3. Check forSymmetricity.
4. Find theAsymptotes.
5. Intervals ofIncrease and Decrease.
6. Evaluate theMaxima and Minimaof the function.
7. ExamineConcavityand the Point of Inflection.
8. Sketch the Graph.

(A) Determine Domain

The given function is \(y = x + \cos x\).

The domain of the given function is \(\mathbb{R}\) as the function is defined everywhere.

(B) Determine Intercepts

Find \(y\)-intercepts by substituting 0 for \(x\) into \(y = x + \cos x\).

\(\begin{array}{c}y = 0 + \cos 0\\ = 1\end{array}\)

So,\(y\)-intercept is \(\left( {0,1} \right)\).

Find \(x\)-intercepts by substituting 0 for \(y\) into \(y = x + \cos x\).

\(\begin{array}{c}0 = x + \cos x\\x \approx - 0.74\end{array}\)

So, \(x\)-intercept is \(\left( { - 0.74,0} \right)\) where \(n\) is an integer.

(C) Determine symmetry

For the given function \(y = x + \cos x\), \(f\left( { - x} \right) \ne x\), \(f\left( { - x} \right) \ne - x\) and \(f\left( {p + x} \right) \ne x\), so the function has no symmetry.

(D) Find Asymptotes 

Determine \(\mathop {\lim }\limits_{x \to \pm \infty } y\) for Horizontal asymptotes.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \pm \infty } y = \mathop {\lim }\limits_{x \to \pm \infty } \left( {x + \cos x} \right)\\ = \pm \infty \end{array}\)

So, there is no horizontal asymptote because the value is not finite.

Determine \(\mathop {\lim }\limits_{x \to {0^ + }} y\) and \(\mathop {\lim }\limits_{x \to {0^ - }} y\) for Vertical asymptotes.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {0^ + }} y = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x + \cos x} \right)\\ \ne \infty \end{array}\)

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {0^ - }} y = \mathop {\lim }\limits_{x \to {0^ - }} \left( {x + \cos x} \right)\\ \ne \infty \end{array}\)

So, there is no Vertical asymptote.

(E) Find Intervals of Increase or Decrease 

Find the first derivative of the given function with respect to \(x\).

\(\begin{array}{c}y' = \frac{d}{{dx}}\left( {x + \cos x} \right)\\ = 1 - \sin x\end{array}\)

For \(y' = 0\), \(x = \frac{1}{2}\left( {4\pi n + \pi } \right)\), where \(n\) is an integer.

Draw a table for the interval of increasing and decreasing.

Interval	\(y'\)	Behaviour of \(y\)
\(\left( { - \infty ,\infty } \right)\)	+	Increasing

(F) Find Local Minimum and Maximum values 

From the obtained table and the condition of local maxima and minima, it can be said that, there is no maximum or minimum values as there is no sign change.

(G) Determine Concavity and points of Inflection 

Find \(y''\).

\(\begin{array}{c}y'' = \frac{d}{{dx}}\left( {1 - \sin x} \right)\\ = - \cos x\end{array}\)

For \(y'' = 0\), \(x = \pm \frac{\pi }{2} + 2\pi n,\frac{{3\pi }}{2} + 2\pi n\).

Draw a table for concavity for different intervals.

Interval	Sign of \(y''\)	Behaviour of \(y\)
\(\left( { - \frac{\pi }{2} + 2n\pi ,\frac{\pi }{2} + 2n\pi } \right)\)	-	Concave downward
\(x = \frac{\pi }{2} + 2n\pi \)	0	Inflection
\(\left( {\frac{\pi }{2} + 2n\pi ,\frac{{3\pi }}{2} + 2n\pi } \right)\)	+	Concave upward

The inflection point will be at \(x = \frac{\pi }{2} + 2n\pi \) on \(y = x\).

And, \(y\left( {\frac{\pi }{2} + n\pi } \right) = \frac{\pi }{2} + n\pi \).

(H) Draw Graph

By using all the obtained information from step 2 to 8, draw the graph of the given function.