Answer :
[tex]f(x)=3x^{2/5}-x^{3/5}[/tex]
We get the derivative : [tex]f'(x)=3*2/5*x^{2/5-1}-3/5*x^{3/5-1}=\dfrac{6}{5x^{3/5}}-\dfrac{3}{5x^{2/5}}[/tex]
f'(x) is negative for x<0.
Let's look at the critical points other than at x=0:
f'(x)=0 <=> 6/5-(3/5)x^(1/5)=0 <=> x^(1/5)=(6/5)*(5/3)=2 hence x=2^5=32
There is a critical point at x=32; for x<=32 f' is positive, for x>=30 f' is negative.
The derivative is positive for positive xs and negative for negative xs, hence f is increasing for 32>=x>0, and decreasing for x<0 and x>32
We get the derivative : [tex]f'(x)=3*2/5*x^{2/5-1}-3/5*x^{3/5-1}=\dfrac{6}{5x^{3/5}}-\dfrac{3}{5x^{2/5}}[/tex]
f'(x) is negative for x<0.
Let's look at the critical points other than at x=0:
f'(x)=0 <=> 6/5-(3/5)x^(1/5)=0 <=> x^(1/5)=(6/5)*(5/3)=2 hence x=2^5=32
There is a critical point at x=32; for x<=32 f' is positive, for x>=30 f' is negative.
The derivative is positive for positive xs and negative for negative xs, hence f is increasing for 32>=x>0, and decreasing for x<0 and x>32
