The answer to this question lies in its first and second derivative.

First, lets find the first and second derivative.

If, #f(x)=(x+3)^(3/2)-6x^3#

using the power rule, the first derivative is:

#f'(x)=(3/2)(x+3)^(1/2)-6*3x^2#

#=(3/2)(x+3)^(1/2)-18x^2#

Using the power rule once more, the second derivative is:

#f''(x)=(3/2)(1/2)(x+3)^(-1/2)-18*2x#

#=3/(4(x+3)^(1/2))-36x#

Now, recall that if, at an #x#-value, #a#,

#f'(x)>0#, #f(x)# is increasing at #x=a#

#f'(x)<0#, #f(x)# is decreasing at #x=a#

Also, recall that if, at an #x#-value, #a#,

#f''(x)>0#, #f(x)# is concave up at #x=a#

#f''(x)<0#, #f(x)# is concave down at #x=a#

So,

#f'(-1)=(3/2)(-1+3)^(1/2)-18(-1)^2#

#=(3sqrt2)/2-18=(3sqrt2-36)/2# which is clearly negative.

And, #f''(-1)=3/(4(-1+3)^(1/2))-36(-1)#

#=3/(4sqrt2)+36# which is clearly positive.

Thus, #f(x)# is decreasing and is concave up at #x=-1#