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Manager  G
Joined: 20 Aug 2017
Posts: 104
What is the minimum value of the function  [#permalink]

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5 00:00

Difficulty:   45% (medium)

Question Stats: 47% (01:06) correct 53% (01:14) wrong based on 53 sessions

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

A. -4
B. 1
C. 0
D. 4
E. 6
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3239
Re: What is the minimum value of the function  [#permalink]

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1

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.

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Intern  B
Joined: 05 Oct 2019
Posts: 10
Re: What is the minimum value of the function  [#permalink]

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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)

Intern  B
Joined: 26 May 2014
Posts: 1
Re: What is the minimum value of the function  [#permalink]

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uchihaitachi wrote:
Q. What is the minimum value of the function $$f(x) = 3x^2 + 6x +4$$?

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

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
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9417
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Re: What is the minimum value of the function  [#permalink]

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uchihaitachi wrote:
Q. What is the minimum value of the function $$f(x) = 3x^2 + 6x +4$$?

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

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.

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