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QUESTION: Given f(x)=2x^4=4x^2+1, identify any inflection points and find the intervals where the function is concave up or down.

I got this question wrong on a test a few weeks ago; I got lost on it right away. I was hoping the correct answer and some basic steps could help me figure out what I did wrong.
Thanks, Clark

ANSWER: I think there is an error in your equation. Shouldn't it be

f(x) = 2x^4 - 4x^2 + 1 ?

---------- FOLLOW-UP ----------

QUESTION: Yes, sorry; that is what the equation should be. The rest of the question looks correct. I appreciate your assistance. Clark

OK. Let's go with

f(x) = 2x^4 - 4x^2 +1.

The keys to these types of problems is to use the fact that

1) the places where the first derivative equals zero, i.e., df/dx = f' = 0, tell you where the extrema are since these are where the slope of the tangent to the function is horizontal, corresponding to a maximum (like a peak) or a minimum (trough of a valley) of the function occur,

2) the 2nd derivative, d^2f/dx^2 = f'', evaluated at these values of x (corresponding to the extrema) tells you whether the function is at a peak (slope decreasing = concave down) or a trough (increasing = concave up).

To summarize

f'(x0) = 0 --> solve to get x0 where f(x0) is max or min

f''(x0) > 0 --> slope is increasing (eg., going from negative to positive)  --> f(x0) at a trough (like the letter "U"): concave up

f''(x0) < 0 --> slope is decreasing (positive to negative) --> f(x0) at a peak (like the symbol ∩): concave down.

Note that f'' is to f' as f' is to f (reminds me of an SAT question!).

I highly recommend plotting the following results on a x-y graph as you go along.

So, now for your particular polynomial (BTW, getting the minus sign correct makes a big difference since then the function and its derivatives are defined for real numbers).

f'(x) = 8x^3 - 8x = 8x(x^2 - 1) = 0  -->  x = 0, 1, -1 are solutions (roots), so this is where the extrema are.

f''(x0) = -8 for x0 = 0  -->  peak :  f(x0=0) = 1  -->  plot this point and sketch a little peak around it
f''(1)   = 16          --> valley : f(1)        = -1  --> sketch a little valley
f''(-1) = 16          --> valley : f(-1)       = -1  --> valley.

From your plot you should be able to connect the dots and see pretty clearly what the function is doing.

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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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