# Thread: Movement Calc Problem, need help fast!!!!

1. ## Movement Calc Problem, need help fast!!!!

Consider the function f(x)=x^2-2ax-3x+a^2+3a, a-1 ≤ x ≤ a+3, a>0
The absolute maximum value is __________
and this occurs at equals _________
The absolute minimum value is _____________
and this occurs at equals ____________

2. ## Ok here you go

$f(x)=x^2-2ax-3x+a^2+3a$...then $f'(x)=2x-2a-3$...find the zeros... $x=\frac{2a+3}{2}$....it is undefined nowhere...and the original funtion is defined everywhere...off to the second derivative test.. $f''(x)=2$...so it is always positive so ... $x=\frac{2a+3}{2}$ is a relative min...now we just need to check the endpoints.. $f(a-1)=3a+1$... $f(a+3)=3a+9$...and $f(\frac{2a+3}{2})=a^2+9a+9$...and even though its a relative min the max occurs at $x=\frac{2a+3}{2}$ and the min at $x=a-1$

3. ## Thanks

Thanks... I had nothing and was no where close to right!