Question

The maximum of the function f(x)=-3x² -2(k-5)+k-9 is equal to the minimum of the function g(x)=x²-2(k-1)x+k+7 Find all such functions f and g (Solve this task without using derivations)​

Answer 1

The answer is clearly done in the image. Go through it.

Answer 2

The solution for the equation is explain as below.

$\begin{array}{l}f(x)=-3 x^{2}-2(k-5) x+k-9 \\=-3\left\{x^{2}+2\left(\frac{k-5}{3}\right) x+\left(\frac{k-5}{3}\right)^{2}\right\}+3\left(\frac{k-5}{3}\right)^{2}+k-9 \\=-3\left\{x+\left(\frac{k-5}{3}\right)\right\}^{2}+3\left(\frac{k-5}{3}\right)^{2}+k-9\end{array}$

$\text { So, Maximum value of } f(x)=3\left(\frac{k-5}{3}\right)^{2}+k-9$

$\begin{array}{l}g(x)=x^{2}-2(k-1) x+k+7 \\=\left\{x^{2}-2(k-1) x+(k-1)^{2}\right\}-(k-1)^{2}+k+7 \\=\{x-(k-1)\}^{2}-(k-1)^{2}+k+7\end{array}$

$\text { So, Minimum value of } \mathrm{g}(\mathrm{x})=-(k-1)^{2}+k+7$

$\text { Maximum value of } \mathrm{f}(\mathrm{x})=\text { Minimum value of } \mathrm{g}(\mathrm{x})$

$\begin{array}{l}3\left(\frac{k-5}{3}\right)^{2}+k-9=-(k-1)^{2}+k+7 \\3\left(\frac{k^{3}-10 k+25}{9}\right)+k-9=-\left(k^{2}-2 k+1\right)+k+7 \\\left(\frac{k^{2}-10 k+25}{3}\right)+k-9=-\left(k^{2}-2 k+1\right)+k+7 \\k^{2}-10 k+25+3 k-27+3\left(k^{2}-2 k+1\right)-3 k-21=0 \\4 k^{2}-16 k-20=0 \\k^{2}-4 k-5=0 \\(k-5)(k+1)=0\end{array}$

$\text { So, } k=5,(-1) \text {. }\\\text { For } k=5 \text {. }$

$\begin{array}{l}f(x)=-3 x^{2}-4 \\g(x)=x^{2}-8 x+12 \\\text { For } k=-1 \\f(x)=-3 x^{2}+12 x-10 \\g(x)=x^{2}-+4 x+6\end{array}$