First thing to do is to find the derivative. That is [tex]f'(x)=3x^2+18x+15[/tex]. Now we need to set the derivative equal to 0 and solve for the critical numbers. [tex]0=3x^2+18x+15[/tex]. Factor out a 3 to make things a bit easier: [tex]0=3(x^2+6x+5)[/tex]. Factor the quadratic inside the parenthesis to get [tex]0=3(x+1)(x+5)[/tex]. Obviously, [tex]3 \neq 0[/tex], but x + 1 = 0 and x = -1, and x + 5 = 0 and x = -5. Those are the critical numbers. We will make a table with the intervals: -∞<x<-5, -5<x<-1, -1<x<∞, pick a point within those boundaries and test each point in the derivative for the sign, positive or negative, that results. Positive the function is increasing on that interval, negative the function is decreasing on that interval. Test -6 in the first interval. f'(-6)=15, so the function is increasing on that interval. Test -2 in the next interval. f'(-2)=-9, so the function is decreasing on that interval. Test 0 in the last interval. f'(0)=15, so the function is increasing on that interval. We know that since the function is increasing from negative infinity to -5, we have a max value there. Find that exact point by subbing -5 into the original function. f(-5)=18; therefore, we have a max value at (-5, 18). We know that since the function is decreasing from -5 to -1, we have a min value there. Find that exact point y subbing -1 into the original function. f(-1)=-14; therefore, we have a min value at (-1, -14). And there you go!
