1. ## Turning Points

How would you find how many turning points if given an equation with out graphing?

ex. f(x) =22x^4 - 4x^3 + 3x^2 - 2x + 2

Edit:

i think i got it. All you need to do is take the highest exponent and subtract 1, giving you the # of turning points for that function.

In this case it would be 4-1 = 3. There fore f(x) has 3 turning points.

2. ## Re: Turning Points

Originally Posted by sakonpure6
How would you find how many turning points if given an equation with out graphing?

ex. f(x) =22x^4 - 4x^3 + 3x^2 - 2x + 2

Edit:

i think i got it. All you need to do is take the highest exponent and subtract 1, giving you the # of turning points for that function.
That is a reasonable approach but the number you get will include points where the slope is increasing, then the slope becomes 0, then increases again, if I remember correctly these are called inflection points, for example x3 ​does this. Also you might have some non-unique turning points, for example x4 has turning points when 4x3=0 and factorizing that shows that the 3 tuning points are all at x=0.

3. ## Re: Turning Points

Originally Posted by sakonpure6
How would you find how many turning points if given an equation with out graphing?

ex. f(x) =22x^4 - 4x^3 + 3x^2 - 2x + 2

Edit:

i think i got it. All you need to do is take the highest exponent and subtract 1, giving you the # of turning points for that function.

In this case it would be 4-1 = 3. There fore f(x) has 3 turning points.
Normally the second derivation is a polynimial its zeros are the x values for the points of inflections. In your problem will have to find what are the real roots of the second derivative equation.

4. ## Re: Turning Points

Originally Posted by sakonpure6
How would you find how many turning points if given an equation with out graphing?

ex. f(x) =22x^4 - 4x^3 + 3x^2 - 2x + 2

Edit:

i think i got it. All you need to do is take the highest exponent and subtract 1, giving you the # of turning points for that function.

In this case it would be 4-1 = 3. There fore f(x) has 3 turning points.
Stationary points occur where the derivative is 0. Since the derivative of a polynomial is a polynomial of one degree less than the original polynomial, the number of turning points can be ANY NUMBER UP TO OR INCLUDING the degree of the derivative.

5. ## Re: Turning Points

The basic $y= x^4$ has degree 4 but only one turning point. A polynomial of degree n has at most n- 1 turning points but can have fewer (including 0 as for $y= x^5$).