Question

for what value of X,8+(x-3)^2 have the least value?​

Answer 1

Answer:

» 8 + (x - 3)²

Will have the least value if (x - 3)² = 0

» (x - 3)² = 0

» x = 3

Therefore, for x = 3, 8 + (x - 3)² will have the least value.

Answer 2

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

We have given the Equation 8+(x-3)²

and we have to find the least value.

Expanding the given term

=>8+(x-3)²

Identity used here :

(a-b)²=a²+b²-2ab

=>8+x²+3²-2(x)(3)

=>8+x²+9-6x

=>x²-6x+8+9

=>x²-6x+17

Now,it is in the form of a quadratic equation.

Here,a= 1 ,b= -6 & c=17

x= -b/2a

x= -(-6)/2

x= 6/2

x=3

Hence,at x= 3 ,8+(x-3)² have the least value.

Alternative Method:

f(x)=8+(x-3)²

f'(x)= 0+2(x-3)

f'(x)= 2x-6

put derivative equals to zero

=>f'(x)=0

=>2x-6=0

=>2x=6

=>x= 6÷2=3

=>x=3

Now, take double derivatives

f'(x)=2x-6

f"(x)= 2

f"(x) is >0 so,it is minimum