If $f(x)=x^2+a x+b$ has minima at $x=3$ whose value is 5 , then the values of $a$ and $b$ are respectively.
If $f(x)=x^2+a x+b$ has minima at $x=3$ whose value is 5 , then the values of $a$ and $b$ are respectively.
-6,-14
-6,14
14,-6
6,14
Solution
$f(x)=x^2+a x+b$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+\mathrm{a}$ and when $\mathrm{f}^{\prime}(\mathrm{x})=0$, we get $\mathrm{x}=\frac{-\mathrm{a}}{2}$
Now $\mathrm{f}^{\prime}(\mathrm{x})=2$ and $2>0$
$\therefore \mathrm{f}(\mathrm{x})$ has minima at $\mathrm{x}=\frac{-\mathrm{a}}{2}=3 \quad \ldots$ [as per given data]
$\therefore \mathrm{a}=-6$
Since Minimum value of $f(x)$ is 5 at $x=3$, we write
$5=(3)^2+(-6)(3)+b \quad \Rightarrow b=14$