Local Extreme Values & points of inflection - PLS HELP?

Question

Determine the nature of the local extremem values and the points of inflection of any three members of the family of functions in the form f(x)=a|x^n-b|+c? (n>1, aeR,beR,ceR and a not= 0, b not= 0), and draw their graphs using the TI-89?

If someone could give me step by step how to do this I would really appreciate it. I have to submit today and have no idea :)

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Jayme H
S Local Extreme Values & points of inflection - PLS HELP?
Determine the nature of the local extremem values and the points of inflection of any three members of the family of functions in the form f(x)=a|x^n-b|+c? (n>1, aeR,beR,ceR and a not= 0, b not= 0), and draw their graphs using the TI-89?

If someone could give me step by step how to do this I would really appreciate it. I have to submit today and have no idea :)

Sorry...the "e" part is set membership symbol thingy...∈

Answer 1

I infer that the aeR etc. means that a, b, and c are elements of the set of real numbers. I also infer that n is an integer from the fact that x is not restricted to nonnegative real numbers but a, b, and c are explicitly restricted to being real numbers.

I can't help much with the details of graphing your family of functions with the TI-89; my experience is mainly with the TI-83, TI-84, and my TI-86. Invoking the necessary functions on the TI-89 is sufficiently different that I have had trouble using that calculator.

However, finding these values by hand isn't as difficult as that. When doing this problem by hand, it is important to break this function up into different domains separated by boundaries in which x^n = b, which is where the argument inside the absolute value function changes signs. At these points, if they exist - which they won't if n is even and b is negative - the derivative will undergo a step discontinuity and the second derivative will not exist. Otherwise, the function can be treated as

f(x) = ax^n - ab + c
f'(x) = anx^(n-1)
f"(x) = an(n-1)x^(n-2) where x^n > b

and

f(x) = -ax^n + ab + c
f'(x) = -anx^(n-1)
f"(x) = -an(n-1)x^(n-2) where x^n < b

Where x^n = b,

f(x) = c
and f'(x) and f"(x) do not exist.

If n is even and b > 0, minima (a > 0) or maxima (a < 0) exist at x = ±(b^(1/n)) but they are "corners" where the derivative abruptly changes sign. The concavity in the function also abruptly changes sign at this point.

If n is even and b < 0, there are no values where the argument inside the absolute value goes nonnegative, so the absolute value signs effectively become ordinary parentheses and these special extrema do not exist.

If n is even, there is a local extremum at x = 0 where the function is infinitely differentiable and does not have any discontinuities in its slope. Furthermore, there will be an even symmetry about the y axis. If a > 0 and b < 0, or a < 0 and b > 0, the extremum at x = 0 will be a maximum. Otherwise it will be a minimum. If a > 0, the curve will open generally upward and will be concave up at the sides. If b > 0, the center section will have a concavity opposite the sides. Of course, for b < 0, there will be no special center section.

If n is odd, there will be a single global "corner" extremum at x = b^(1/n), which will be a minimum if a > 0 and a maximum if a < 0 and no other extrema. There will, however, be a critical point of inflection at x = 0. Between x = 0 and x = b^(1/n), the concavity of the curve will be opposite that of the curve as a whole, changing abruptly at x = b^(1/n) but going smoothly through zero at x = 0.

Answer 2

I would if you could explain what aeR, beR, and ceR mean.