We can see from f ′ ( x ) = 9 7 ⋅ x + 1 x 5 7 {\displaystyle f'(x)={\frac {9}{7}}\cdot {\frac {x+1}{x^{\frac {5}{7}}}}} that f ′ ( x ) = 0 {\displaystyle f'(x)=0} if x = − 1 {\displaystyle x=-1} and undefined if x = 0 {\displaystyle x=0} . We also know f ( x ) {\displaystyle f(x)} is increasing for x ∈ ( − ∞ , − 1 ) {\displaystyle x\in (-\infty ,-1)} and decreasing for x ∈ ( − 1 , 0 ) {\displaystyle x\in (-1,0)} . Therefore, a local maximum occurs at x = − 1 {\displaystyle x=-1} . Moreover, a local minimum occurs at x = 0 {\displaystyle x=0} since f ( x ) {\displaystyle f(x)} is decreasing for x ∈ ( − 1 , 0 ) {\displaystyle x\in (-1,0)} and increasing for for x ∈ ( 0 , ∞ ) {\displaystyle x\in (0,\infty )} . Hence, f ( x ) has a local minimum at ( 0 , 0 ) and a local maximum at ( − 1 , 7 2 ) {\displaystyle \color {blue}f(x){\text{ has a local minimum at }}(0,0){\text{ and a local maximum at }}\left(-1,{\frac {7}{2}}\right)} .
that 
if 
and undefined if 
. We also know 
is increasing for 
and decreasing for 
. Therefore, a local maximum occurs at . Moreover, a local minimum occurs at since is decreasing for and increasing for for 
. Hence, 
.
