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Let f(x) = 5x^(1/3) - x^(5/3) Find the intervals of concavity and the inflections point(s). Graph...

Let f(x) = 5x^(1/3) - x^(5/3)

Find the intervals of concavity and the inflections point(s).

Graph the function.

Solutions

Expert Solution

for stationary point we set f"(x)=0

=> x=0

At x=0 f(x)=f(0)=0-0=0

Therefore stationary point is (0,0)

For concave upward f"(x)>0

=> x^(1/3)<0

f(x) is concave up on

for concave downward f"(x)<0

f is concave down on

The graph looks like

milcah answered 1 year ago

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