Answer
(a) Increasing: $(\frac{1}{4}, \infty )$
(b) Decreasing :$( -\infty, \frac{1}{4} )$ except $x=0$
(c) Concave up:$ (−∞,−\frac{1}{2})∪(0,∞)$
(d) Concave down:$ (−\frac{1}{2},0),$
(e) Inflection points: $x=−\frac{1}{2},0$
Work Step by Step
(a) Intervals where f is increasing.
$f(x)=x^{\frac{4}{3}}-x^{ \frac{1}{3}}$
$f'(x)=\frac{d}{dx}f(x)=\frac{d}{dx}x^{\frac{4}{3}}-\frac{d}{dx}x^{ \frac{1}{3}}$
$f'(x)= \frac{4}{3}x^{\frac{4}{3}-1}- \frac{1}{3} x^{\frac{1}{3}-1}=\frac{4}{3}x^{\frac{1}{3}}- \frac{1}{3} x^{-\frac{2}{3}} $
$f'(x)=\frac{1}{3}(4x^{\frac{1}{3}} -\frac{1}{x^{\frac{2}{3}}})= \frac{1}{3} \frac{4x^{\frac{1}{3}} x^{\frac{2}{3}}-1}{x^{ \frac{2}{3}}}= \frac{1}{3}( \frac{4x-1}{x^{\frac{2}{3}}} )$
$f'(x)=\frac{1}{3x^{\frac{2}{3}}}(4x-1) $
For $x\ne0 $, $ \frac{1}{3x^{\frac{2}{3}}}\gt0 $, So
$f'(x) \gt 0 \Longleftrightarrow 4x-1\gt 0 $
$\Longleftrightarrow 4x\gt 1 \Longleftrightarrow x\gt\frac{1}{4} $
Hence
f(x) is increasing on $( \frac{1}{4}, \infty)$
(b)Intervals where f is decreasing
$f'(x)\lt 0 \Longleftrightarrow 4x-1\lt 0 $
That is $x\lt \frac{1}{4}$
f(x) is decreasing on $(-\infty, \frac{1}{4})$
(c) Open intervals where f is concave up.
$f'(x)=\frac{4}{3}x^{\frac{1}{3}} -\frac{1}{3}x^{- \frac{2}{3}}$
$f''(x)=\frac{d}{dx}f'(x) =\frac{d}{dx}\frac{4}{3}x^{\frac{1}{3}} - \frac{d}{dx}\frac{1}{3}x^{- \frac{2}{3}}$
$f''(x)=\frac{d}{dx}f'(x) =\frac{4}{3} \frac{d}{dx} x^{\frac{1}{3}} - \frac{1}{3} \frac{d}{dx} x^{- \frac{2}{3}}$
$f''(x)=\frac{d}{dx}f'(x) =\frac{4}{3} \frac{1}{3} x^{\frac{1}{3}-1} - \frac{1}{3} (-\frac{2}{3}) x^{- \frac{2}{3}-1}$
$f''(x)=\frac{4}{9}x^{-\frac{2}{3}}+\frac{2}{9}x^{-\frac{5}{3}} $
$f''(x)=\frac{2}{9}x^{-\frac{5}{3}} (2x)+\frac{2}{9}x^{-\frac{5}{3} }=\frac{2}{9} x^{-\frac{5}{3}}(2x+1)$
Now
For $x\gt 0 , x^{-\frac{5}{3} } \gt 0, $ and $2x+1\gt 0, So f''(x)\gt 0$
For $ -\frac{1}{2}\lt x \lt 0: $ , $ x^{-\frac{5}{3}}\lt 0$ and $2x+1\gt 0,$ So $f''(x)\lt 0$
For $x\lt-\frac{1}{2}: x^{-\frac{5}{3}}\lt0 $ and $2x+1\lt 0$, $f''(x)\gt 0$
Therefore, concave up on $(-\infty,-\frac{1}{2}) $ and on $(0, \infty)$
Concave up on $(-\infty,-\frac{1}{2}) \cup (0, \infty) $
(d) Open intervals where 𝑓 is concave down
Concave down on $(-\frac{1}{2},0) $
(e) x-coordinates of inflection points
Inflection points occur where concavity changes sign. From the intervals, concavity changes
at $x=-\frac{1}{2}$ and at $x=0$.Both points are in the domain of f.
Their y-coordinates are
$f(-\frac{1}{2})= (-\frac{1}{2} )^{ \frac{4}{3}} - (-\frac{1}{2} )^{ \frac{1}{3}} = (\frac{1}{2} )^{ \frac{4}{3}} + (\frac{1}{2} )^{ \frac{1}{3}} , f(0)=0 $
Inflection points at $x= -\frac{1}{2}, $ and $x=0$
