What are the extrema of # f(x)=(x^2 -9)^3 +10# on the interval [-1,3]?

2 Answers
Mar 31, 2016

Answer:

We have a minima at #x=0# and a point of inflection at #x=3#

Explanation:

A maxima is a high point to which a function rises and then falls again. As such the slope of the tangent or the value of derivative at that point will be zero.

Further, as the tangents to the left of maxima will be sloping upwards, then flattening and then sloping downwards, slope of the tangent will be continuously decreasing, i.e. the value of second derivative would be negative.

A minima on the other hand is a low point to which a function falls and then rises again. As such the tangent or the value of derivative at minima too will be zero.

But, as the tangents to the left of minima will be sloping downwards, then flattening and then sloping upwards, slope of the tangent will be continuously increasing or the value of second derivative would be positive.

If second derivative is zero we have a point of

However, these maxima and minima may either be universal i.e. maxima or minima for the entire range or may be localized, i.e. maxima or minima in a limited range.

Let us see this with reference to the function described in the question and for this let us first differentiate #f(x)=(x^2-9)^3+10#.

Its first derivative is given by #f'(x)=3(x^2-9)^2*2x#

= #6x(x^4-18x^2+81)=6x^5-108x^3+486x#.

This would be zero for #x^2-9=0# or #x=+-3# or #0#. Of these only #{0,3}# are within the range #[-1,3}#.

Hence maxima or minima occurs at points #x=0# and #x=3#.

To find whether it is maxima or minima, let us look at second differential which is #f''(x)=30x^4-324x^2+486# and hence while

at #x=0#, #f''(x)=486# and is positive

at #x=3#, #f''(x)=2430-2916+486=0# and is a point of inflection.

Hence, we have a local minima at #x=0# and a point of inflection at #x=3#

. graph{(x^2-9)^3+10 [-5, 5, -892, 891]}

Mar 31, 2016

Answer:

The absolute minimum is #(-9)^3+10# (which occurs at #0#), the absolute maximum on the interval is #10#, (which occurs at #3#)

Explanation:

The question does not specify whether we are to find relative or absolute extrema, so we will find both.

Relative extrema can occur only at critical numbers. Critical numbers are values of #x# that are in the domain of #f# and at which either #f'(x)=0# or #f'(x) does not exist. (Fermat's Theorem)

Absolute extrema on a closed interval can occur at critical numbers in the interval or at enpoints of the interval.

Because the function asked about here is continuous on #[-1,3]#, the Extreme Value Theorem assures us that #f# must have both an absolute minimum and absolute maximum on the interval.

Critical numbers and relative extrema.

For #f(x) = (x^2-9)^3+10#, we find #f'(x) = 6x(x^2-9)^2#.

Clearly, #f'# never fails to exist, so there are no critical numbers of that kind.

Solving #6x(x^2-9)^2 = 0# yields solutions #-3#, #0#, and #3#.

#-3# is not in the domain of this problem, #[-1,3]# so we need only check #f(0)# and #f(3)#

For #x < 0#, we have #f'(x) < 0# and

for #x > 0#, we have #f'(x) > 0#.

So, by the first derivative test, #f(0)# is a relative minimum. #f(0) =-9^3+10#.

The other critical number in the interval is #3#. If we ignore the domain restriction, we find that #f'(x) > 0# for all #x# near #3#. So, the function increases on small open intervals containing #3#. Therefore, if we stop at #3# we have hit the highest point in the domain.

There is not universal agreement whether to say that #f(3)=10# is a relative maximum for this function on #[-1,3]#.

Some require value on both sides to be less, others require values in the domain on either side to be less.

Absolute Extrema

The situation for absolute extrema on a closed interval #[a,b]# is much simpler.

Find critical numbers in the closed interval. Call the #c_1, c_2# and so on.

Calculate the values #f(a), f(b), f(c_1), f(c_2)# and so on. The greatest value is the absolute maixmum on the interval and the least value is the absolute minimum on the interval.

In this question we calculate #f(-1) = (-8)^3 +10#, #f(-3) = 10# and #f(0) = (-9)^3+10#.

The minimum is #f(0) = (-9)^3+10# and

the maximum is #f(-3) = 10#.