Suppose $f(x)$ is a function satisfying the following conditions \(i) $f(0)=2, f(1)=1$ \(ii) $f$ h

Question

Suppose $f(x)$ is a function satisfying the following conditions \(i) $f(0)=2, f(1)=1$ \(ii) $f$ has a minimum value at $x=5 / 2$, and \(iii) for all $x$, $f^{\prime}(x)=\left|\begin{array}{ccc}2 a x & 2 a x-1 & 2 a x+b+1 \ b & b+1 & -1 \ 2(a x+b) & 2 a x+2 b+1 & 2 a x+b\end{array}\right|$ \where $a, b$ are some constants. Determine the constants $a$ and $b$

Professor PC Thomas Classes · Answer

Applying $R_{3} ightarrow R_{3}-R_{1}-2 R_{2}$, we get\ $f^{\prime}(x)=\left|\begin{array}{ccc}2 a x & 2 a x-1 & 2 a x+b+1 \ b & b+1 & -1 \ 0 & 0 & 1\end{array}ight|$\ $=\left|\begin{array}{cc}2 a x & 2 a x-1 \ b & b+1\end{array}ight|$\ $=\left|\begin{array}{cc}2 a x & -1 \ b & 1\end{array}ight| \quad$ (using $C_{2} ightarrow C_{2}-C_{1}$ )\ or $f^{\prime}(x)=2 a x+b$\ Integrating we get $f(x)=a x^{2}+b x+c$\ where $c$ is an arbitrary constant.\ Since $f$ has a maximum at $x=5 / 2$, therefore\ $f^{\prime}\left(\frac{5}{2}ight)=0$ or $5 a+b=0$.\ Also $f(0)=2 \quad \Rightarrow c=2$.\ and $f(1)=1 \quad \Rightarrow a+b+c=1$\ $	herefore \quad a+b=-1$\ Solving (1) and (2), we get\ $a=1 / 4$, and $b=-5 / 4$\ $	herefore f(x)=\frac{1}{4} x^{2}-\frac{5 x}{4}+2$\