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

Question

William Abdalla · Accepted Answer

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

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

Emily Ammerman · Answer

undefined To find the absolute extrema of $ f(x) = x^5 - x^3 + x^2 - 7x $ in the interval $[0, 7]$, first, evaluate the function at the critical points and endpoints within the interval. Then, compare the function values to determine the maximum and minimum.

1. Find the critical points by setting the derivative equal to zero and solving for $ x $.
2. Evaluate $ f(x) $ at the critical points and endpoints within the interval.
3. Identify the maximum and minimum values among these function values.

The critical points occur where the derivative is zero or undefined. Taking the derivative of $ f(x) $ yields $ f'(x) = 5x^4 - 3x^2 + 2x - 7 $. To find critical points, set $ f'(x) = 0 $ and solve for $ x $. Then evaluate $ f(x) $ at these points and the endpoints of the interval $[0, 7]$.

Finally, determine which of these function values is the maximum and minimum within the interval.