F(x)=2x^3-3x^2-12x+k
a)show that if k=30, then f(x)=0 has at least one solution
b)show that if k=30, then f(x)=0 has exactly one solution
c)find all values of k so that f(x)=0 has exactly one solution

I'm going to replace f(x) with y since we're talking about graphing here. 

dy/dx = 12 - 6x - 6x^2Set this = 0 

-6x^2 - 6x + 12 = 0Divide both sides by (-6) 

x^2 + x - 2 = 0 

(x + 2) * (x - 1) = 0 

x = 1, -2 

Take the second derivative: d^2y/dx^2 = -6 - 12x 

This is negative at x = 1 and positive at x = -2 so we have a maximum at x = 1 and a relative minimum at x = -2 

The function is increasing on (-2, 1) < Answer for (a) 
 

f(x)= k +12x - 3x^2 - 2x^3 

f(1) = k + 12 - 3 - 2 = k + 7 

If you want that to be 4, then k + 7 = 4 and k = - 3 < Answer for k---(b) 
 

I guess that I should be considering k to be -3 from here on out. You weren't specific about whether I should leave it as k or change it to -3. 

So from now on the equation is: f(x)= -2x^3 -3x^2 + 12x - 3 < 
 

f(-2) = 16 -12 -24 - 3 = 16 - 39 = - 23 

So the minimum is (-2, -23) < Answer to (c) 

 

(-∞,-2) < Answer to (d)