Let f(x): x^3 - 6x^2 + 9x
When x= ..?, the graph of y=f(x) has a min. value

Question

Let f(x): x^3 - 6x^2 + 9x 
When x= ..?, the graph of y=f(x) has a min. value

campbell_st · Answer

find the 1st derivative and solve for x... 

you will get 2 values

then find the 2nd derivative

test the 1st derivative solutions in the 2nd derivative to determine their nature, max. min or horizontal point of inflexion. 

hope it helps

campbell_st · Answer

an alternate method is to graph the curve here is a link https://www.desmos.com/calculator

anonymous · Answer

Like this?
f'(x) = 0 
3x² − 12x + 9 = 0 
3 (x² − 4x + 3) = 0 
3 (x − 1) (x − 3) = 0 
x = 1, x = 3 

f''(1) = 6(1) − 12 = −6 < 0
f''(3) = 6(3) − 12 = 6 > 0

But then the answer would be 3, while on the answer sheet the right answer should be 1

campbell_st · Answer

so the minimum is at x = 3 since f"(3) > 0

and max at x = 1 since f"(1) < 0

your solution is correct

anonymous · Answer

Yeah, but the correct answer should be 1.

Also may I ask why we need to find the 2nd derivative too? Is it only to use the x value we got from the 1st derivative?

campbell_st · Answer

did you graph the curve?

if you graph it you'll see the relative max and mins

did you enter f(x) correctly.... are you sure its not 

f(x) = -x^3 - 6x^2 + 9x

anonymous · Answer

Ah, I'm sorry, I got the question wrong (should be the maximum value). Thank you so much for the help!!