Inflection Points: Understanding the Concept and Identifying Them in a Function

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Homework Help Overview

The discussion revolves around identifying inflection points and understanding concavity for the function f(x) = -8x^4 - 5x^3 + 3. Participants explore the relationship between the second derivative and the concept of inflection points.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of inflection points and their relation to the second derivative. Questions arise about the conditions under which inflection points occur, specifically regarding the need for the second derivative to change sign.

Discussion Status

There is an ongoing exploration of the conditions for concavity and inflection points. Some participants have provided corrections and clarifications regarding the second derivative and its roots, while others are attempting to confirm their understanding of concavity intervals.

Contextual Notes

Participants are working under the constraints of a homework assignment, which includes identifying intervals of concavity based on the second derivative. There is a noted confusion regarding the terminology used for concavity and the implications of the second derivative's sign.

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Homework Statement


f(x)=-8x4-5x3 +3 for concavity and inflection points.

Homework Equations


f(x)=-8x4-5x3 +3
f'(x)=-32x3-15x2
f''(x)=-96x2-30x

The Attempt at a Solution


can someone explain to me what exactly Inflection Points are?
is it like critical points for the first derivative but those are for the second one?
i mean, making f''(x) = 0 and get 2 numbers and those are the Inflection Points?
 
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Indeed. Inflection points are where the function changes from being concave-up to concave-down or vice versa. This is the same as the local extrema of the derivative, or the roots of the second derivative.
 
LumenPlacidum said:
Indeed. Inflection points are where the function changes from being concave-up to concave-down or vice versa. This is the same as the local extrema of the derivative, or the roots of the second derivative.

so i would do this
f''(x)=-96x2-30x
f''(x)=-3x(32x-10)

x=0, 10/32

and those are my answers? well i tried and it doesn't work.. :(
 
f''(x) = -96x2 - 30x = -6x(16x + 5)
So f''(x) = 0 for x = 0 and x = -5/16
 
LumenPlacidum said:
Indeed. Inflection points are where the function changes from being concave-up to concave-down or vice versa. This is the same as the local extrema of the derivative, or the roots of the second derivative.

No, it isn't quite the same. The roots of the second derivative are the values of x where there may but there need not be an inflection point. You must always additionally check that the second derivative actually changes sign at those roots. For example, if

f''(x) = (x-2)(x-5)^2

then x = 2 gives location for an inflection point but x = 5 does not.
 
Mark44 said:
f''(x) = -96x2 - 30x = -6x(16x + 5)
So f''(x) = 0 for x = 0 and x = -5/16

oh yeah i didnt notice the - sign mistake :S

there is a second part to this question which asks to find the points where it concaves up and down.
the graph of f is only concaving up isn't it?
 
The graph of f is concave up (concaving isn't a word) when f''(x) > 0, and is concave down when f''(x) < 0. For your problem, f''(x) is positive for some x values and negative for other x values.
 
this is how the question is constructed:
"f is concave down on (-infty ,___)U(___, infty) and its concave up on (___,___)"

since f''(x) = 0 for x = 0 and x = -5/16
so when you said "f''(x) is positive for some x values and negative for other x values"
==> x = 0, -5/16

so would it be:
"f is concave down on (-infty ,-5/16)U(0, infty) and its concave up on (-5/16,0)"
?
 
  • #10
Mark44 said:
Works for me.

sweet... thanks for your explanation!
 

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