Answer: stationary point \((-2,0)\), minimum.

We are given

\(\mathrm f''(x)=(2x+5)^{-\frac32}.\)

Integrate once to find \(\mathrm f'(x)\):

\(\mathrm f'(x)=\int(2x+5)^{-\frac32}\,\mathrm dx.\)

Since \(\dfrac{\mathrm d}{\mathrm dx}(2x+5)=2\),

\(\int(2x+5)^{-\frac32}\,\mathrm dx=-(2x+5)^{-\frac12}+c.\)

So

\(\mathrm f'(x)=1-\dfrac1{\sqrt{2x+5}}\)

after finding the constant as follows. The gradient is \(\dfrac23\) when \(x=2\), so

\(-9^{-\frac12}+c=\dfrac23.\)

Thus

\(-\dfrac13+c=\dfrac23,\)

giving \(c=1\).

At a stationary point, \(\mathrm f'(x)=0\), so

\(1-\dfrac1{\sqrt{2x+5}}=0.\)

Hence

\(\sqrt{2x+5}=1,\)

so

\(2x+5=1,\qquad x=-2.\)

Now integrate \(\mathrm f'(x)\) to find \(\mathrm f(x)\):

\(\mathrm f(x)=\int\left(1-(2x+5)^{-\frac12}\right)\,\mathrm dx.\)

This gives

\(\mathrm f(x)=x-\sqrt{2x+5}+d.\)

The curve passes through \((2,2)\), so

\(2=2-\sqrt9+d.\)

Thus \(d=3\), and

\(\mathrm f(x)=x-\sqrt{2x+5}+3.\)

At \(x=-2\),

\(\mathrm f(-2)=-2-1+3=0.\)

Therefore the stationary point is

\((-2,0).\)

To determine the nature, use the second derivative:

\(\mathrm f''(-2)=(2(-2)+5)^{-\frac32}=1\gt0.\)

So the stationary point is a minimum.