Question

Describe the end behavior, the maximum number of x-intercepts, the existence of a maximum or minimum value for each of the following functions. Use graphing technology to confirm your thinking.

Answer 1

2, a)

Given:

$f\mleft(x\mright)=-3x^3+3x^2-2x+1$

Aim:

We need to find the end behavior, the maximum number of x-intercepts, the existence of a maximum or minimum value of the given functions.

Explanation:

Use graphing technology.

The graph of the given function is

One end of the curve is moving upward to infinity when x tends to negative infinity and another end is moving down to negative infinity when x tends to infinity.

Take the limit to infinity on the given function to find the end behavior.

$\lim_{x\to\infty}f\mleft(x\mright)=\operatorname{\lim}_{x\to\infty}\mleft(-3x^3+3x^2-2x+1\mright)$

$\lim_{x\to\infty}f\mleft(x\mright)=-3\infty+3\infty-2\infty+1=-\infty$

$\lim_{x\to\infty}f\mleft(x\mright)=-\infty$

Take limit to negative infinity on the given function to find the end behavior.

$\lim_{x\to-\infty}f\mleft(x\mright)=\operatorname{\lim}_{x\to-\infty}\mleft(-3x^3+3x^2-2x+1\mright)$

$\lim_{x\to-\infty}f\mleft(x\mright)=\infty$

End behavior:

$f\mleft(x\mright)=\infty\text{ as x}\rightarrow-\infty$

$f\mleft(x\mright)=-\infty\text{ as x}\rightarrow\infty$

We know that the x-intercept is the intersection point where the function f(x) crosses the x-axis.

The x-intercepts = (0.718.0)

Differentiate the given function with respect to x and set the result to zero.

Solve for x to find the existence of a maximum or minimum value of the given function.

$f\mleft(x\mright)=-3x^3+3x^2-2x+1$

Differentiate the given function with respect to x

$f^{\prime}\mleft(x\mright)=-3\times3\left(x^2\right)+3\times2\left(x\right)-2$

$f^{\prime}\mleft(x\mright)=-9x^2+6x-2$

SEt f'(x) =0 and solve for x.

$-9x^2+6x-2=0$

Multiply both sides by (-1).

$9x^2-6x+2=0$

which is of the form

$ax^2+bx+c=0$

where a =9, b=-6 and c=2.

Use quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute a =9, b=-6, and c=2 in the equation.

$x=\frac{-\lparen-6)\pm\sqrt{\left(-6\right)^2-4\times9\times2}}{2\times9}$

$x=\frac{6\pm\sqrt{36-72}}{18}$

$x=\frac{6\pm i6}{18}$

$x=\frac{1+i}{2},\frac{1-\imaginaryI}{2}$

We get a complex value for x.

There is no maximum or minimum value of the given function in a real number.

Final answer:

$f\mleft(x\mright)=\infty\text{ as x}\rightarrow-\infty$

$f\mleft(x\mright)=-\infty\text{ as x}\rightarrow\infty$

The x-intercepts = (0.718.0)

There is no maximum or minimum value for the given equation.