Finding Points of Inflection for a Given Function: A Homework Challenge

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In summary: So, to find y intercepts for f(x), we need to find the second derivative at those points and use the rule for finding y intercepts for second derivatives.
  • #1
Dr Zoidburg
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Homework Statement


Suppose the graph f''(x) of a function is given by:
(see attachment)

(a) Find all points of inflection of f(x)

The Attempt at a Solution


First I figured, by looking at the graph and seeing the intercept points [(-2,0),(0,0),(2,0)] that f''(x) = [tex]x^{4}-4x^{2}[/tex]
solving for f''(x)=0 gives us x = -2 or 0 or 2.
Between -2 < x < 2 there is no change of sign, which indicates no point of inflection.
Thus the points of inflection for f(x) are at x = -2 and x = 2.
correct?
But how can I find the y intercepts with just what I have? I only know how to find them by putting x back into f(x). As we don't have f(x) here, I'm stuck. Integrating back doesn't help because of the unknown variables.
Or am I trying too hard, and what I've done is the answer?
 

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  • #2
Well, remember that an inflection point of f(x) is where the first derivative is at either a maximum or minimum. And find first the maxima and minima of the first derivative, we must solve the zeros of the third derivative =D
 
  • #3
Dr Zoidburg said:

Homework Statement


Suppose the graph f''(x) of a function is given by:
(see attachment)

(a) Find all points of inflection of f(x)

The Attempt at a Solution


First I figured, by looking at the graph and seeing the intercept points [(-2,0),(0,0),(2,0)] that f''(x) = [tex]x^{4}-4x^{2}[/tex]
Okay, that looks good.

solving for f''(x)=0 gives us x = -2 or 0 or 2.
?? If f(x)= x4 - 4x2 then f '(x)= 4x3- 8x and f "(x)= 12x2- 8. That is not 0 at x= 0, 2, and -2!

Between -2 < x < 2 there is no change of sign, which indicates no point of inflection.
Thus the points of inflection for f(x) are at x = -2 and x = 2.
correct?
But how can I find the y intercepts with just what I have? I only know how to find them by putting x back into f(x). As we don't have f(x) here, I'm stuck. Integrating back doesn't help because of the unknown variables.
Or am I trying too hard, and what I've done is the answer?
 
  • #4
third derivative is [tex]4x^{3}-8x[/tex]
solving for f'''(x) = 0 gives us
x = +/-[tex]\sqrt{2}[/tex] and x = 0.

minima points of f''(x) are at x = (-[tex]\sqrt{2}[/tex], -4) & ([tex]\sqrt{2}[/tex], -4)
maxima point of f''(x) is (0, 0).

I still confused here! How does this help me find the y co-ordinates for inflection points of f(x)?
 
  • #5
HallsofIvy said:
?? If f(x)= x4 - 4x2 then f '(x)= 4x3- 8x and f "(x)= 12x2- 8. That is not 0 at x= 0, 2, and -2!
the graph x4 - 4x2 is the second derivative, not f(x).
 

What is a point of inflection?

A point of inflection is a point on a curve where the direction of the curve changes from concave up to concave down, or vice versa. This means that the slope of the curve changes sign at the point of inflection.

How do you find points of inflection?

To find points of inflection, you need to take the second derivative of the function and set it equal to zero. Then, solve for the x-value(s) that make the second derivative equal to zero. These x-values are the points of inflection.

Why are points of inflection important?

Points of inflection can provide valuable information about the behavior of a curve. They can help determine the minimum and maximum values of a function, as well as the points of inflection can also help identify the presence of critical points, where the first derivative is equal to zero.

Can a point of inflection be a local or global extremum?

No, a point of inflection cannot be a local or global extremum. Extrema occur at points where the first derivative is equal to zero, while points of inflection occur at points where the second derivative is equal to zero.

Are points of inflection always present in a function?

No, points of inflection are not always present in a function. A function may have zero, one, or multiple points of inflection, depending on the behavior of the curve. Some functions may not have any points of inflection at all.

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