Question 262269
here is the problem:
(i) {{{f(x)= 5+15x+6x^2-x^3}}}
to determine increasing and decreasing, take a second derivative as
(ii) {{{f'(x) = 15 + 12x - 3x^2}}}
set = 0 and find x. We get
(iii) {{{f'(x) = 15 + 12x - 3x^2 = 0 }}}
divide by -3 to get
(iv) {{{x^2 - 4x - 5 = 0}}}
(v) {{{(x+1)(x-5) = 0}}}
so, x = -1 and x = 5.
these are critical values.
From (ii) take a second derivative as
(vi) f''(x) = {{{12 - 6x = 0}}}
and solving for x, we get
{{{x = 2}}}
this is also a critical value.
pick a number less than and greater than 2 in to f'' and we get
x= 0 - - > increasing ; x = 4 - - > decreasing.
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so the intervals of increasing and decreasing are
increasing (-00, 2)
decreasing (2, +00)
here is the graph to help.

{{{ graph( 300, 200, -10, 10, -20, 100, 5+15x+6x^2-x^3) }}}