how many turning points?

Question

x^2 (x-2) (x^2 + 3)

is there 4 turning points in this function? thanks.

Answer 1

No, just 2. When x < 0, all 3 factors are < 0 and the graph is in quadrant 3. The function has a double zero at x=0, so there the graph just touches the x axis and turns away. There's a zero at x=2, so between 0 and 2 the graph turns again and crosses the x axis at x=2. The factor x²+3 produces no real zeros, so the graph just continues up to the right after x=2.

Answer 2

While there could be 4 turning points in a 5th degree equation, this one only has 2 turning points.

Answer 3

at most 4
The degree is 5, so the maximum number of turning points is
5-1 = 4
For this function, there are only 2.

Answer 4

The derivative of f(x) is

f'(x)=x(5x^3-8x^2+9x-10).

The zeros of f'(x) (exactly) are

x=0, x =(8 - 71/(2942 + 45*Sqrt[4451])^(1/3) +
(2942 + 45*Sqrt[4451])^(1/3))/15,
x=8/15 + (71*(1 + I*Sqrt[3]))/(30*(2942 + 45*Sqrt[4451])^(1/3)) -
((1 - I*Sqrt[3])*(2942 + 45*Sqrt[4451])^(1/3))/30},
{x -> 8/15 + (71*(1 - I*Sqrt[3]))/(30*(2942 + 45*Sqrt[4451])^(1/3)) -
((1 + I*Sqrt[3])*(2942 + 45*Sqrt[4451])^(1/3))/30

which are approximately

x=0, x=1.4796830201032358, x=0.06015848994838202 + 1.272143845602291*I, x=0.06015848994838202 - 1.272143845602291*I

In this case, (0,0) is a relative maximum and (approximately) (1.47,-5.91) is a relative minimum.

Answer 5

there are 2 turnin oints
bcuz x is only square and not cub not to the 4th power.
it go up from -infinity to x=0 and go down to x=1.5 &y= -6 then go up passing by x=2
so it turn down and up

Answer 6

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Answer 7

yes