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#
