# h(x)=(x+1)^2-3 What is the minimum value of h(x)?

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We have the function h(x)=(x+1)^2-3. We need to find the minimum value of h(x). To do this we fund the first derivative of h(x) and equate it to zero.

h(x) = (x+1)^2-3

h'(x) = 2(x + 1)

equating it to zero and solving for x

=> 2( x + 1) =0

=> x = -1

At x = -1, h(x) = (-1 + 1)^2 - 3 = -3.

Also h''(x) = 2 which is positive for x = -3. This means we have found the minimum value.

**Therefore the minimum value of the function is -3.**

h(x) = (x+1)^2-3. To find the minimum.

Since (x+1)^2 > = 0, for all x, as (x+1)^2 is a square.

(x+1)^2 - 3 > = 0 -3.

Theefore h(x) > = -3 for all x.

Therefore minimum of h(x) = -3, when (x+1) = 0, or when x= -1.

or h(-1) = -3 is the minimum of h(x) .