Question

given the f(x)=-x^3+12x^2+60x-50 help asap medal!!!!!

over what interval is f(x) increasing?

Answer 1

find f'(x)
then find the values for which f'(x)>0

Answer 2

can you help me more with it @surjithayer

Answer 3

did you get the derivative?

Answer 4

yea for the first one but not the second one f'(x) = -3x^2 + 24x + 60

Answer 5

Given: f(x)=-x^3+12x^2+60x-50
Find the first and second derivativ es.
Set the first derivative = to 0 and solve for x. Should obtain 2 roots. These are called the "zeros" or "roots" or "solutions" of f(x)=-x^3+12x^2+60x-50.

Answer 6

but I'm i suck on the second one;/

Answer 7

Next, plot these two roots on the x-axis. Form three subintervals based upon these roots. For example, \[(-\infty , a), (a, b), (b, \infty)\]

Answer 8

i got =-0x

Answer 9

Choose a test number from each interval. Check the sign of the first derivative at each such number. If the deriv. is +, the function is increasing on that interval.

It's difficult to respond to "i got =-0x." Please, share all of y our work.

Answer 10

-3 times 2 x ^2-1
-3 times 2
-6 x^1 -6 ^2-1
-12x^1
-12x
-3x
-0x

Answer 11

Your result, "f'(x) = -3x^2 + 24x + 60," is fine. You must set this = to zero and solve for the 2 roots. Could you do that now, please?

Hint: 3 is a common factor of f'(x) = -3x^2 + 24x + 60. Take it out.

Answer 12

Hint: one root of this equation is 10. Can you demonstrate that 10 is a root? What is the other root?

Answer 13

f'(x) = -3(x^2 - 8x - 20)=0. x=? x=?

Answer 14

Note how I factored "-3" out of each term within parentheses, above?