The graph of the function has one relative maximum and one relative minimum poin

Question

The graph of the function has one relative maximum and one relative minimum point. Plot these two points and check the concavity there. Using only this information, sketch the graph. f(x)x3x-5x-12 The relative minimum point on the graph is (Simplify your answer. Type an ordered pair. Use integers or fractions for any numbers in the expression.) Since the value of f" at this relative minimum point is at this point, the graph is The relative maximum point on the graph is (Simplify your answer. Type an ordered pair. Use integers or fractions for any numbers in the expression.) Since the value of f" at this relative maximum point isat this point, the graph is

Explanation / Answer

f'(x)=-x^2+6x-5=0

gives, x=1,5

f''(x)=-2x+6

f''(1)=4

f''(5)=-4

Hence,

Since, f''(1)>4 hence it is point of miniimum

Since, f''(5)<-4 hence it is point of maximum

f(5)=-125/3+3*25-5*5-12=-11/3

Hence relative minimum point is (5,-11/3)

f(1)=-1/3+3-5-12=-1/3-14=-43/3

So relative maximum point is

(1,-43/3)

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