# find the minimmum value for f(x) = x^2 + 6x -5

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### 6 Answers

f(x) = x^2 + 6x - 5

First we need to obtain the first derivative's zeros:

f'(x) = 2x + 6 = 0

==> 2x = -6

==> x= -3

Now we have a critical values at x= -3 which means the the function f(x) has a minimmum values at f(-3)

==> f(-3) = (-3)^2 + 6*-3 - 5

= 9 - 18 - 5 = -14

Then the minnimum value is:

**f(-3) = -14 OR the point (-3,-14)**

The minimum value of the function f(x) = x^2 + 6x -5 has to be determined.

The function f(x) = x^2 + 6x - 5 can be rewritten in the following way as the sum of a square and a constant.

x^2 + 6x - 5

= x^2 + 6x + 9 - 5 - 9

= (x + 3)^2 - 14

The minimum value of (x +3)^2 is 0 as the square of a real number is positive.

The polynomial (x+3)^2 - 14 can take on a minimum value of -14.

When x = -3, f(x) = -14, this is the minimum value of the function.

To calculate the local extremes of a function, minimum or maximum, we'll do the first derivative test.

f'(x) = (x^2 + 6x -5)'

f'(x) = 2x + 6

Now, we'll calculate the roots of the first derivative. Each root of derivative represents the value for the function f(x) has an extreme value.

f'(x) = 0

2x + 6 = 0

We'll factorize by 2:

2(x+3) = 0

We'll divide by 2:

x + 3 = 0

We'll subtract 3 both sides:

x = -3

Now, we'll calculate the minimum value of the function:

f(-3) = (-3)^2 + 6*(-3) - 5

f(-3) = 9 - 18 - 5

f(-3) = -14

**The coordinates of the minimum point are: (-3 , -14).**

Minimum and maximum value for any function f(x) occurs for value of x that corresponds to value of f'(x) equal to 0.

The given function is:

f(x) = x^2 + 6x - 5

For this:

f'(x) = 2x + 6

Equating the above expression for f'(x) to 0:

2x + 6 = 0

==> 2x = - 6

==> x = - 3

Substituting this value of x in expression for f(x)

f(-3) = (-3)^2 + 6*(-3) - 5

= 9 -18 - 5 = -14

Answer:

Minimum value of f(x) = -14

f(x) = x^2+6x-5

To find the minimum of f(x).

Consider x^2+6x-5 = x^2+6x+3^2 -14 ,as -5 is replaced by 3^2-14.

x^2+6x+5 = (x^2+6x+3^2) - 14

x^2+6x+5 = (x+3)^2 - 14 .

When x= -3 , RHS (x+3)^2 - 14 = (-3+3)-14 is least as (x+3)^2> 0 for all other values of x except x = -3.

So when x =-3,

x^2+6x+5 = (x+3)^2-14 = (-3+3)^2 - 14 = 0-14 has the least value.

So f(x) > -14 for all x < 0 and x > 0 and f(-3) = -14 is minnimum.

We have the function f(x) = x^2 + 6x - 5.

Diffrentiating it with respect to x, we use the relation: if f(x) = x^n, f'(x) = nx^(n-1).

f'(x)= 2x + 6

Now equate 2x + 6 =0

x= -6 / 2 = -3

Therefore the function has the minimum value at x = -3

At x= -3

y = ( -3) ^2 + 6 ( -3 ) -5

= 9 - 18 - 5

= -14

**Therefore the minimum value is - 14.**