Find the critical numbers of f(x)=3x^4−36x^3−1512x^2.

a) Roots

b) y-intercept

c) Inflection points in (x,y) form

d) Maximums in (x,y) form

e) Minimums in (x,y) form

To find the critical numbers, set the derivative equal to 0 and solve for x:
f(x) = 3x^4 - 36x^3 - 1512x²
f '(x) = 12x^3 - 108x² - 3024x
12x^3 - 108x² - 3024x = 0
12x(x-21)(x+12) = 0
12x = 0 or x - 21 = 0 or x + 12 = 0
x = 0 or x = 21 or x = -12
So the critical numbers are 0, 21, and -12.

a) To find the roots of the function, set the function equal to 0 and solve for x:
f(x) = 3x^4 - 36x^3 - 1512x²
3x^4 - 36x^3 - 1512x² = 0
3x²(x² - 12x - 504) = 0
3x² = 0 or x² - 12x - 504 = 0
x² = 0 or x² - 12x = 504
x = 0 or x² - 12x + 36 = 504 + 36
x = 0 or (x - 6)² = 540
x = 0 or x - 6 = ±√540
x = 0 or x = 6 ±√540
x = 0 or x = 6 ± 6√15
So the roots are x = 0, x = 6 + √15, and x = 6 - √15.

b) To find the y-intercept, just set x equal to 0 and solve for f(x):
f(x) = 3x^4 - 36x^3 - 1512x²
f(x) = 3(0)^4 - 36(0)^3 - 1512(0)²
f(x) = 0
So the y-intercept is f(x) = 0.

c) To find the inflection points, find the second derivative, set it equal to 0, and solve for x, then plug that back into f(x) to find the y-coordinate:
f(x) = 3x^4 - 36x^3 - 1512x²
f '(x) = 12x^3 - 108x² - 3024x
f ''(x) = 36x² - 216x - 3024
36x² - 216x - 3024 = 0
36(x² - 6x - 84) = 0
36 = 0 (impossible) or x² - 6x - 84 = 0
x² - 6x = 84
x² - 6x + 9 = 84 + 9
(x - 3)² = 93
x - 3 = ±√93
x = 3 ±√93
So the x-coordinates of the points of inflection are 3 + √93 and 3 - √93. To find the y-coordinates, just plug those values back into f(x) (it's not difficult, but it's messy, so I won't do it here)

d) and e) To find the relative maximums and minimums, use the first derivative to test a point between each of the critical numbers that we found in the beginning (and ±∞):
f '(-13) = 12(-13)^3 - 108(-13)² - 3024(-13) = -5304