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