Sketching the graph and finding the function by using the given. A function f that is defined and continuous for $x\neq 0$ satisfies the following conditions:


*

*f($2-{\sqrt 2}$) = $\sqrt[3]{1-1/\sqrt 2}$, f($\frac{2}{3}$) = $0$,  f(2) = $\sqrt[3]{2}$, f($2+{\sqrt 2}$) = $\sqrt[3]{1+1/\sqrt 2}$

*$\lim_{x \to 0^-}f(x)$ = $-\infty$, $\lim_{x \to 0^+}f(x)$ = $\infty$, $\lim_{x \to -\infty}f(x)$ = $0$, $\lim_{x \to \infty}f(x)$ = $0$

*$f^\prime$($x$) $<$ $0$ for $x$ $<$ $\frac{2}{3}$ and $x\neq 0$, and for $x$ $>$ $2$; $f^\prime$($x$) $>$ $0$ for $\frac{2}{3}$ $<$ x $<$ $2$

*$\lim_{x \to \frac{2}{3}^-}f^\prime(x)$ = $ -\infty$,  $\lim_{x \to \frac{2}{3}^+}f^\prime(x)$ = $\infty$ 

*$f^{\prime\prime}(x)$ $<$ $0$ for $x$ $<$ $0$ and for $2-{\sqrt 2}$ $<$ $x$ $<$ $2+{\sqrt 2}$ and $x\neq \frac{2}{3}$;$f^{\prime\prime}(x)$ $>$ $0$ for $0$ $<$ $x$ $<$ $2-{\sqrt 2}$ and for $x$ $>$ $2+{\sqrt 2}$
The question asks to come up with function $f$($x$)$ = $ (a$x$ $+$ b)$^{c}$$x$$^{d}$ that satisfies 1-5 conditions. What are a, b, c and d?
I drew the graphand found that d is obviously -1 however I couldn't find a, b, and c. Please help. 
 A: First, since the left and right limits at zero are unequal (in fact, do not exist), $f$ is discontinuous at $0$, so that $d<0$. Since $f\left(\frac{2}{3}\right)=0$, the factor $(ax+b)^c$ must vanish there, so that $c>0$ and $ax+b = r(3x-2)$ for some constant $r$. So rewriting, we get
$$f(x) = r^c(3x-2)^c x^d.$$
Next, $f(2)>0$ shows that $r^c>0$; assume going forward that in fact $r=1$, and we will try to fix things up later if that proves not to work. This gives
$$f(x) = (3x-2)^c x^d.$$
Then $f(2) = 2^{2c+d} = 2^{1/3}$, so that $2c+d = \frac{1}{3}$. Next,
\begin{align*}
f(2-\sqrt{2})f(2+\sqrt{2}) &= ((4-3\sqrt{2})(4+3\sqrt{2}))^c((2-\sqrt{2})(2+\sqrt{2}))^d \\
&= (-2)^c2^{d} = (-1)^c2^{c+d}\\
\left(1-\frac{1}{\sqrt{2}}\right)^{1/3}\left(1-\frac{1}{\sqrt{2}}\right)^{1/3}
&= \left(\frac{1}{2}\right)^{1/3} = 2^{-1/3}.
\end{align*}
Since these two expressions must be equal, we see that $(-1)^c=1$ and $c+d = -\frac{1}{3}$.
Solving
\begin{align*}
  2c+d &= \phantom{-}\frac{1}{3}\\
\phantom{2}c+d &= -\frac{1}{3}
\end{align*}
gives $c=\frac{2}{3}$ and $d=-1$, so that
$$f(x) = (2x-3)^{2/3}x^{-1}.$$
Although we did not use the various conditions on the derivatives, this function is easily seen to satisfy all of them.
A plot is shown below:

