Question

Use the guidelines of section 4.4 to sketch the curve

15.\(y = {x^4} - 3{x^3} + 3{x^2} - x\)

Short answer

The graph of the given function is drawn.

Step-by-step solution

Domain of \(y\)

The domain of a polynomial is all the real numbers.

Thus, the domain of the polynomial \(y = {x^4} - 3{x^3} + 3{x^2} - x\) is \(\left( { - \infty ,\infty } \right)\).

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,\(y\left( 0 \right) = 0\)implies y-intercept is 0. Similarly, let y=0 then,

\(\begin{aligned}{c}{x^4} - 3{x^3} + 3{x^2} - x = 0\\x\left( {{x^3} - 3{x^2} + 3x - 1} \right) = 0\\x{\left( {x - 1} \right)^3} = 0\end{aligned}\)

Thus, \(x = 0,1\) are the x-intercepts.

Symmetry of \(y\)

Consider the given function as\(f\left( x \right) = {x^4} - 3{x^3} + 3{x^2} - x\).

Substitute x by –x then,

\(\begin{aligned}{c}f\left( { - x} \right) = {\left( { - x} \right)^4} - 3{\left( { - x} \right)^3} + 3{\left( { - x} \right)^2} - \left( { - x} \right)\\ = {x^4} + 3{x^3} + 3x + x\end{aligned}\)

The above polynomial is not equal to\(f\left( x \right)\)or\( - f\left( x \right)\).

Thus, the given function is neither odd nor even.

Asymptotes of \(y\)

Since, the function is finite for all values of x implies there is no vertical asymptote.

And there is no horizontal asymptote since,\(\mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right) = \infty \).

Thus, there is no vertical and horizontal asymptote.

Local maximum and minimum

The first derivative of the given function is\(f'\left( x \right) = 4{x^3} - 9{x^2} + 6x - 1\). Equate\(f'\left( x \right) = 0\)and find x:

\(\begin{aligned}{c}4{x^3} - 9{x^2} + 6x - 1 = 0\\{\left( {x - 1} \right)^2}\left( {4x - 1} \right) = 0\end{aligned}\)

\(x \in \left( { - \infty ,\frac{{ - 1}}{4}} \right)\)For ,\(f'\left( x \right) < 0\)implies\(f\left( x \right)\)is\(\left( { - \infty ,\frac{{ - 1}}{4}} \right)\)on the interval . For\(x \in \left( {\frac{{ - 1}}{4},\infty } \right)\),\(f'\left( x \right) > 0\)implies\(f\left( x \right)\)is increasing.

\(x > 1\)Thus, the local minimum is at \(x = \frac{{ - 1}}{4}\).

Concavity and inflection point

Find the second derivative of the given function.

\(\begin{aligned}{c}f''\left( x \right) = 12{x^2} - 18x + 6\\ = 6\left( {2{x^2} - 3x + 1} \right)\\ = 6\left( {2x - 1} \right)\left( {x - 1} \right)\end{aligned}\)

Equate\(f''\left( x \right) = 0\)then,\(x = 1,\frac{1}{2}\). Thus, the inflection point is at\(x = 1,\frac{1}{2}\).

The graph is concave up in the interval \(\left( {\frac{1}{2},1} \right)\) and concave down at \(x < \frac{1}{2}\) and .

Graph of \(y\)

Draw the graph of the given function as shown below:

Therefore, the graph of the given function is drawn.