Question

Determine the nature of all critical points of f(x)=-3x^4+8x^3, then sketch a rough graph of the function. Thank you for the help!!

Answer 1

slope = -12 x^3 + 24x^2
curvature = -36 x^2 + 48 x
so we need to sketch a graph
Where is y = 0?
y = x^3 (-3x +8)
when x = 0 and when x = 8/3
what happens when |x| is big ?
y = x^3 (-3x +8) ===> -3 x^4
well x^4 is a big positive number way left or way right of the origin
so y dives for - infinity way left and way right
Now where is the slope = 0 ?
slope = -12 x^3 + 24x^2 = x*2 (-12 x + 24) = 12 x^2 (-x+2)
That is zero at zero and at x = 2
so what is y when x = 2?
y = x^3 (-3x +8) ===> 8 (-6+8) = 16
so the function is horizontal at the origin and at (2,16)
is it a maximum at (2,16) ?
at x = 2 curvature = -36 x^2 + 48 x = -36*4+48*2 = -48
so that curves down at (2,16)
at the origin the curvature is zero, so it just pauses
now start sketching

Answer 2

Critical points occur where the derivative of the function is equal to zero. In this case, the first derivative of f(x)=-3x^4+8x^3 is -12x^3+24x^2 which can be factored into (-12)(x^2)(x-2). This implies that our critical points when f(x)=0 are located at x=0, and x=2. If we plug in x=0 into the function, we see that our first critical point is (0,0) and if we plug in x=2 into the function, then our second critical point is (2,16). Therefore, your critical points for the function f(x)=-3x^4+8x^3 are (0,0) and (2,16).