Question

What are the absolute extrema of \[f(x)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}}\] on \[-1,1\]?

Options

• Absolute maximum value \[\frac{-9}{4}\] at \[x=\frac{1}{8}\], and absolute minimum value \[18\] at \[x=-1\].

• Absolute maximum value \[0\] at \[x=0\], and absolute minimum value \[6\] at \[x=1\].

• Absolute maximum value \[18\] at \[x=-1\], and absolute minimum value \[\frac{-9}{4}\] at \[x=\frac{1}{8}\].

• Absolute maximum value \[6\] at \[x=1\], and absolute minimum value \[0\] at \[x=0\].

Answer 1

The values at the endpoints and critical points are \[18,0,-9/4,6\]. Comparing these values shows that \[18\] is greatest and \[-9/4\] is least.