g(x)=-3x²+30x-71

a. does function have a minimum or a maximum value

b. what is the function's minimum or maximum value

c. where does it occur x=

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the function has a maximum and minimum value at g' = 0

the derivative of g(x) will be

g' = 6x+ 30

a) since g' exists then the function has maximum and minimum values

b)

at maximum or minimum

g' = 0

6x + 30 = 0

6x= -30

x= -5

= 3(-5)^2 +30(-5) -71

= 75 - 150 - 71

g(x)= -146

c. Occurs at x=-5

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