What are the absolute extrema of # f(x)= x^5 -x^3+x^2-7x in [0,7]#?

1 Answer
Jul 2, 2017

Answer:

Minimum: #f(x) = -6.237# at #x= 1.147#

Maximum: #f(x) = 16464# at #x = 7#

Explanation:

We're asked to find the global minimum and maximum values for a function in a given range.

To do so, we need to find the critical points of the solution, which can be done by taking the first derivative and solving for #x#:

#f'(x) = 5x^4 - 3x^2 + 2x - 7#

#x ~~ 1.147#

which happens to be the only critical point.

To find the global extrema, we need to find the value of #f(x)# at #x=0#, #x = 1.147#, and #x=7#, according to the given range:

  • #x = 0#: #f(x) = 0#

  • #x = 1.147#: #f(x) = -6.237#

  • #x = 7#: #f(x) = 16464#

Thus the absolute extrema of this function on the interval #x in [0, 7]# is

Minimum: #f(x) = -6.237# at #x = 1.147#

Maximum: #f(x) = 16464# at #x = 7#