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.

Step-by-step 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)=-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)=\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*3\left(x^2\right)+3*2\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=(-b\pm√(b^2-4ac))/(2a)`

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

`x=(-\lparen-6)\pm√(\left(-6\right)^2-4*9*2))/(2*9)`

`x=(6\pm√(36-72))/(18)`

`x=(6\pm i6)/(18)`

`x=(1+i)/(2),(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.