Question

a)State the minimum value of (x-2)^2 +1 and the corresponding value of x. b)Sketch the graph of the function f(x)= (x-2)^2 +1 for domain: -1<x<5 and write down the range of f corresponding to the domain​

Answer 1

Applying AM > GM

(x²+1/x²)/2 > (x²+1/x²)^1/5

(x²+1/x²) > 2

hence the minimum value is 2

OR

so,

2^x = 2/x^3;

x^4 = 1;

it will give u four roots, two will be imaginary numbers becoz they work in pair form so except those two there are two more solutions available here and those are the real numbers so;

Taking real solutions, x = 1 or -1.

Using second derivative,

d(2^x − 2 /x^3)/ dx = 2 + 6 x^4

At x = +1 or -1, the value is positive, hence minima occurs.....

so put the value into the equation and the result is .....

1+1/1 = 2 , which is the minimum value for the equation

Answer 2

${\huge \underline{\mathfrak\green{hy \:mate}}}$

${\huge{\bf\red{answer :}}}$

is writing an answer

RockstarPratheek

RockstarPratheekMaths AryaBhatta

Applying AM > GM

(x²+1/x²)/2 > (x²+1/x²)^1/5

(x²+1/x²) > 2

hence the minimum value is 2

OR

so,

2^x = 2/x^3;

x^4 = 1;

it will give u four roots, two will be imaginary numbers becoz they work in pair form so except those two there are two more solutions available here and those are the real numbers so;

Taking real solutions, x = 1 or -1.

Using second derivative,

d(2^x − 2 /x^3)/ dx = 2 + 6 x^4

At x = +1 or -1, the value is positive, hence minima occurs.....

so put the value into the equation and the result is .....

1+1/1 = 2 , which is the minimum value for the equatio

${\huge{\bf\red{Thanks,}}}$

I hope it's helpful