What is the minimum value of the function

Question

What is the minimum value of the function  f(x) = 3x^2 + 6x +4 ?

A. -4
B. 1
C. 0
D. 4
E. 6

EgmatQuantExpert · Answer

Solution

Given  
In this question, we are given that
 • A function  f(x) = 3x^2 + 6x + 4

To find  
We need to determine
 • The minimum value of f(x)

Approach and Working out  
We can write f(x) as:
 •  f(x) = 3x^2 + 6x + 4 
Or,  f(x) = 3(x^2 + 2x + 1) + 1 
Or,  f(x) = 3(x + 1)^2 + 1

Now, for f(x) to be minimum,  3(x + 1)^2  has to be minimum
 • Minimum value of  3(x + 1)^2 = 0 
• Therefore, minimum value of f(x) = 0 + 1 = 1

Thus, option B is the correct answer.

Correct Answer: Option B

nerdynerdnerd · Answer

By observation, there are only 2 possibilities to give the smallest value; (i) x = 0; (ii) x = -ve integer.

Substituting x = 0, you get 4
Substituting x = -1, you get 1
Substituting x = -2, you get 4 (you will know by now the more -ve you go for x, the bigger the value)

Hence the answer is 1

MDUsman · Answer

To get the Minimum value for the equation the Differentiation of the Equation must be equated to 0.So,
df(x) d(3x^2 +6x+4)
____ = 0 ==> _____________ = 0 ; 
dx dx

6x+6 = 0
 x= -1;

Putting x= -1 in equation gives the ans (3-6+4) = 1

ScottTargetTestPrep · Answer

Solution:

We see that the function is a quadratic function with a positive coefficient of x^2. Thus, its graph is an up-opening parabola with its vertex as its lowest point. Therefore, the y-value of the vertex is the minimum value of the function.

The x-value of the vertex can be obtained using the formula: x = -b/(2a), where a and b are the coefficients of x^2 and x, respectively. Therefore, the x-value of the vertex is -6/(2(3)) = -1, and the y-value of the vertex, i.e., the minimum value of the function, is f(-1) = 3(-1)^2 + 6(-1) + 4 = 3 - 6 + 4 = 1.

Alternate Solution:

Alternatively, we can re-express the given quadratic function in vertex form, i.e. in the form f(x) = a(x - h)^2 + k. In this form, the coordinates of the vertex will be given by (h, k), and the minimum value of the function will be k. Setting 3x^2 + 6x + 4 equal to a(x - h)^2 + k, we obtain:

a(x - h)^2 + k = 3x^2 + 6x + 4

a(x^2 - 2xh + h^2) + k = 3x^2 + 6x + 4

ax^2 - 2ahx +ah^2 + k = 3x^2 + 6x + 4

Equating the coefficients of the quadratic terms, we get a = 3. Equating the coefficients of the linear terms and substituting a = 3, we get:

-2(3)hx = 6x

-6h = 6
h = -1

Finally, equating the constant terms, we obtain:

ah^2 + k = 4

3(-1)^2 + k = 4

3 + k = 4

k = 1

So, the vertex is (-1, 1), and the minimum value of the function is 1.

Answer: B

1happyArish · Answer

why not f(0) = 4 is the minimum?
If we take f(1) = 13 (it is greater)
and other values like :
f(-1) = 1
f(-2) = 4
are greater and keep increasing if we increase the -ve values

i see that the 4 minimum 
Help me understand

Bunuel · Answer

f(-1) gives 1, which is LESS than f(0) = 4.

What is the minimum value of the function

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What is the minimum value of the function

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