What are the global and local extrema of #f(x) = 2x^3 - 5x +1# ?

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Nimo N. Share
Feb 9, 2018

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Please see below. A graph is also included.

Explanation:

Problem:
What are the global and local extrema of
# color(blue)( f(x) = 2x^3 − 5x + 1 # ?

  • The terms "local extrema" and "relative extrema" are often used interchangeably, however they can be used in specific and different ways, depending somewhat on the person using the terms.

"local extrema" is often associated with finding extreme values over a specific, well defined open interval, where as the term "relative extrema" usually refers to extreme values found by using 1st and 2nd derivatives, regardless of the actual domain of the function.

Assuming, the derivative method is intended for finding such values {and, I shall use the term "relative extrema"}:
# color(red)( f'(x) = 6x^2 − 5 #, and
# color(red)( f''(x) = 12x #

Setting # f'(x) = 0 # will yield critical points, to be explored further to determine their nature.
# f'(x) = 6x^2 − 5 = 0 #
# 6x^2 = 5 #
# x^2 = 5/6 #
# x = +- sqrt(5/6) = +-(sqrt(30))/6 #
# color(red)( x ~~ +- 0.9129 #, critical points

Evaluate # f''(x) # at x = 0 and for the above values to determine the nature of the cps.

# color(red)( f''(0) = 0 #, so x = 0 is an inflection point.

# color(red)( f''(0.9129) = 12*(0.9129) #. The exact value isn't needed. Since this is positive, the curve "holds water" here, so this location is a relative minimum value.
# color(red)( f(sqrt(5/6)) ~~ -2.0429 #

# color(red)( f''(-0.9129) = 12*(-0.9129) #. Since this is negative, the curve "spills water" here, so this location is a relative maximum value.
# color(red)( f(-sqrt(5/6)) ~~ 4.0429 #

Global max and min?
Since,
# Lim_(x rarr oo) f(x) = oo #, and
# Lim_(x rarr -oo) f(x) = -oo #,
the global max and min are at positive and negative infinity, respectively.

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