If $\left(\mathrm{\alpha },\mathrm{\beta }\right)$ and $(\mathrm{\gamma },\mathrm{\delta })$ where $\mathrm{\alpha }<\mathrm{\delta }$, are the turning points of $\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^2-15{\mathrm{x}}^2+36\mathrm{x}-8$ then $\mathrm{\alpha }-\mathrm{\gamma }-\mathrm{\beta }+\mathrm{\delta }=$

$$\begin{gathered} {\text{⨍ (x)=2x}}^3{\text{-15x}}^2\text{+36x-8} \\ {\text{⨍}}^{ 1}{\text{(x)=6x}}^2\text{-30x+36} \\ =6({\mathrm{x}}^2-5\mathrm{x}+6) \\ =6(\mathrm{x}-2) (\mathrm{x}-3) \end{gathered}$$

$\text{⨍ (x)}$ is min or max

$$\begin{gathered} \Rightarrow {\text{⨍}}^{ 1}\text{(x)=0} \\ \mathrm{x}=2,3 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=2 \text{⨍ (x)=16-60+72-8} \\ =20 \\ \mathrm{x}=3 \text{⨍ (x)=54-135+108-8} \\ =19 \\ \therefore (2, 20) (3, 19) \end{gathered}$$

are turning points

$$\begin{gathered} \mathrm{\alpha }=2 \mathrm{\beta }=20 \mathrm{\gamma }=3 \mathrm{\delta }=19 \\ \mathrm{\alpha }-\mathrm{\gamma }-\mathrm{\beta }+\mathrm{\delta }=2-3-20+19=-2 \end{gathered}$$