Question

for all the real values of x, the minimum value of |x+2|-1 is?​

Answer 1

Answer:

Let's find the first derivative of this function

(x

2

+x+1)

2

(2x−1)(x

2

+x+1)−(2x+1)(x

2

−x+1)

Putting this to zero, we get

x

2

−1=0

Which gives us x=±1

Since we can see that the double derivative is quite difficult to solve, so lets put the value of x in main function and see the result.

So for x=1 the function is

3

1

and for x=−1 function is 2

So minimum value is

3

1

at x=1

Step-by-step explanation:

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