I am in college algebra and am confused on how to find absolute maximum and minimum on the graphing calculator. the equation is f(x)=x^(2/3)*(x-4)

Question

anonymous · Answer

$$f \left( x \right)=x ^{\frac{ 2 }{ 3 }}\left( x-4 \right)=x ^{\frac{ 5 }{ 3 }}-4x ^{\frac{ 2 }{ 3 }}$$
$$f \prime \left( x \right)=\frac{ 5 }{3 }x ^{\frac{ 2 }{3 }}-4*\frac{ 2 }{3 }x ^{\frac{ -1 }{ 3 }}$$
$$f \prime \left( x \right)=\frac{ 5x-8 }{3x ^{\frac{ 1 }{3 }} }$$
$$f \prime \left( x \right)=0~ gives $$
5x-8=0,
$$x=\frac{ 8 }{5 }$$
$$f \prime \prime \left( x \right)=\frac{ 5 }{ 3 }*\frac{ 2 }{ 3 }x ^{\frac{ -1 }{ 3 }}-\frac{ 8 }{ 3  }*\frac{ -1 }{  3  }x ^{\frac{ -4 }{  3 }}$$
$$f \prime \prime \left( x \right)=\frac{ 10 }{ 9x ^{\frac{ 1 }{ 3 }} }+\frac{ 8 }{ 9x ^{\frac{ 4 }{ 3  } } } $$

at x=8/5 find f"(x)

anonymous · Answer

$$f \prime \prime \left( x \right)=\frac{ 10x+8 }{ 9x ^{\frac{ 4 }{ 3 }} }$$
f"(x)>0 at x=8/5
there is minima at x=8/5 ,there is no maxima at x=8/5
put x=8/5 in f(x) and find absolute minima.