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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.

Describe the end behavior, the maximum number of x-intercepts, the existence of a-example-1
User Alocaly
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1 Answer

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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)=\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*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.

Describe the end behavior, the maximum number of x-intercepts, the existence of a-example-1
User Vmr
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