The function [tex]f(x) = 2x^3 - 45x^2 + 300x + 10[/tex] has one local minimum and one local maximum.

This function has a local minimum at [tex]x = \_\_\_\_\_\_\_[/tex] with function value [tex]\_\_\_\_\_\_\_\_\_\_[/tex] and a local maximum at [tex]x = \_\_\_\_\_\_\_[/tex] with function value [tex]\_\_\_\_\_\_\_\_\_\_[/tex].

A) [tex]x = 7.5[/tex], minimum value = 1.25, maximum value = 265

B) [tex]x = 15[/tex], minimum value = 5, maximum value = 310

C) [tex]x = 7.5[/tex], minimum value = 265, maximum value = 1.25

D) [tex]x = 15[/tex], minimum value = 310, maximum value = 5

To find local and global extrema of a function on a given interval, calculate the critical points by taking the derivative and setting it equal to zero. In this case, the function f(x)=x²-4x+5 has no local extrema within the interval [-1,1].

Explanation:

The function f(x)=x²-4x+5 can be defined on the interval [-1,1]. To find any local and global extrema, first find the critical points by taking the derivative: f'(x)=2x-4. Setting the derivative equal to zero gives the critical point at x=2. However, upon further analysis, this point is a point of inflection, not a maximum or minimum, so the function has no local extrema.