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
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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

\]

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