d Problem: For x=4 the function f(x)=x^2+bx+c has the minimum value of -9, find ‘c’. | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Problem: For x=4 the function f(x)=x^2+bx+c has the minimum value of -9, find ‘c’.

Problem: For x=4 the function ​$$f(x)=x^2+bx+c$$​ has the minimum value of -9, find ‘c’.

Solution:

We take one time derivative of the function ;

$f'(x)=2x+b$

Now put x=4, we get,

$f'(4)=2(4)+b=8+b$

Since f(x) has minimum value at x=4, its 1st derivative at x=4 should be equale to zero i-e,

$f'(4)=8+b=0$

or

$b+8=0$

or

$b=-8$

Now it is given that at x=4 the function f(x) has minimum value as -9, it means,

$f(4)=4^2+b(4)+c=-9$

or

$16+4b+c=-9$

put value of b=-8, we get

$16-32+c=-9$

$c=7 So the value of c is 7$