# Stationary Points of Inflection

1. May 31, 2004

### Cummings

Now, given y=x^3 -9x^2+23x-16 on the interval [-3,7]

the maximum and minimum values would be the turning points right?

also, a stationary point of inflextion is where the grandient is zero, with a positive or negative gradient on both sides right?

i am asked to find the EXACT values of the x-coordinates of the points of inflection on the graph of -x^4 + 3x^3 + 5x^2 + 2x + 11

but, there are two maximums and one minimum, not a stationary point of inflection.

So, are minimum and maximum values points of inflections?

Also, how do i obtain an EXACT value for the x coordinates.

2. May 31, 2004

### arildno

You are mixing together the concepts "point of inflection" and "saddle point"

A "saddle point" is a point where the derivative is zero, but where the function does not achieve a local extremal value.

A "point of inflection" is a point where the curvature changes sign; in particular, the 2.derivative is zero at a "point of inflection".

3. May 31, 2004

### e(ho0n3

Since you're given an interval, then you're dealing with local extrema here. I'm not sure what you mean by 'turning point'. I suggest you take a look at the definition of a local max. and min. again.

As arildno said, you're probably getting some of these concepts mixed up. There is a so-called second derivative test which you can use to find whether a given point on the function is a a local maximum or minimum. It basically says:

If for some p, f'(p) = 0 then
- If f''(p) > 0, there is a local minimum at (p, f(p))
- If f''(p) < 0, there is a local maximum at (p, f(p))
- If f''(p) = 0, then you're out of luck.

Hope that helps,
e(ho0n3

4. May 31, 2004

### confuted

Be careful with that. A point of inflection only occurs if the 2nd derivative changes signs, not simply if it is zero. For example, if $$f(x)=3x$$, then $$f''(x)=0$$ for all values of x, but there are clearly no inflection points.

But if you look at $$f(x)=x^3$$, $$f''(x)=6x$$ and we see that $$f''(0)=0, f''(-.1)=-.6, f''(.1)=.6$$ and the 2nd derivative is clearly changing signs at x=0, so there is an inflection point there.

5. May 31, 2004

### arildno

Hmm..that's what I meant with ("a point where the curvature changes sign"), but I see now that the last part of the sentence made the meaning ambiguous.
Thx.