Question

Minimum value of the function

f {x} =x^2+x+1
is ....................

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Answer 1

Answer:

let f(x)=y

y=x^2+x+1

dy/dx=2x+1

for minimum value of x dy/dx must be equal to zero

dy/dx=0

2x+1=0

x = -1/2

put value of x in f(x)

f(-1/2) = (-1/2)^2+(-1/2)+1

1//4-1/2+1

3/4

minimum value of x is 3/4