Answer to Question #147656 in Calculus for liam donohue

Question

Consider the function f(x)=−2x3+45x2−300x+5. For this function there are three important intervals: (−∞,A], [A,B], and [B,∞) where A and B are the critical values.
Find A

and B

For each of the following intervals, tell whether f(x) is increasing (type in INC) or decreasing (type in DEC).
(−∞,A]:

[A,B]:

[B,∞)

f(x) has an inflection point at x=C
where C is

Finally for each of the following intervals, tell whether f(x) is concave up (type in CU) or concave down (type in CD).
(−∞,C]:

[C,∞)

Answer 1

f(x) = -2x³+45x²-300x+5

f'(x) = -6x²+90x-300

= -6(x²-15x+50)

= -6(x²-10x-5x+50)

= -6(x-5)(x-10)

For critical point f'(x)= 0

So -6(x-5)(x-10) =0

=> (x-5)(x-10)= 0

=> x = 5 or 10

So A = 5 and B = 10

Case-1:

Let x ≤ 5

Obviously x < 10

So (x-5) ≤0 and (x-10) < 0

=> (x-5)(x-10)≥0

=> -6(x-5)(x-10)≤0

=> f'(x) ≤0

So f(x) is decreasing for x≤5

Case-2:

Let 5≤x≤10

So (x-5) ≥0 and (x-10) ≤ 0

=> (x-5)(x-10)≤0

=> -6(x-5)(x-10)≥0

=> f'(x) ≥0

So f(x) is is increasing for 5≤x≤10

Case-3:

Let x ≥ 10

Obviously x > 5

So (x-5) > 0 and (x-10) ≥ 0

=> (x-5)(x-10)≥0

=> -6(x-5)(x-10)≤0

=> f'(x) ≤ 0

So f(x) is decreasing for x ≥ 10

f''(x) = -12x+90

Now for critical point f''(x) = 0

=> -12x+90 = 0

=> 12x = 90

=> x = 90/12 = 15/2

=> x = 15/2

A curve of a function y = f(x) is concave up where f''(x) > 0 and concave down where f''(x) < 0

If x > 15/2 , f''(x) = -12(x-15/2) < 0

If x < 15/2 , f''(x) = -12(x-15/2) > 0

So f(x) is concave up in interval (-∞, 15/2) concave down in interval (15/2, ∞)

So the answers are given by

A = 5 and B = 10

(-∞,5] : f(x) is decreasing

[5,10] : f(x) is increasing

[10,∞): f(x) is decreasing

C = 15/2 = 7.5

(-∞, 7.5) : concave up

(7.5, ∞) : concave down