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If 'x' is maximum at x = 1, then \(\frac{d}{dx}f(x) = 0\) at x = 1
\(\frac{d}{dx}f(x) = 2ax + b = 0\) Therefore, putting value of x =1 we get \(b = -2a\)
Now going back to \(f(1) = 3\), we get \(a + b + 1 = 3\) --> \(a - 2a + 1 = 3\)
Thus, \(a = -2\) and \(b = 4\)
So our quadratic function is : \(f(x) = -2x^2 + 4x + 1\)
Thus \(f(10) = -159\)
Answer : A
Ps. I doubt this is a GMAT problem. Please mention the source.
_________________
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A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is
a)-159, b)-110, c)-180, d)-105, e)-119
Pls explain
Not sure this would be on the GMAT, seems you should use calculus for this.
ax^2+bx+c=0
f(1)=3=a+b+c f(0)=1=c
f'(x)= 2xa+b.
since the max is 3, f'(1)=0. Thus, 0=2a+b--> b=-2a --> 2=a-2a --> -2=a
f(10) = -2*100+10(4)+1 --> -159
This can be solved in the following way too:
We have \(f(x)=ax^2+bx+c\).
\(f(0)=c=1\) --> \(f(x)=ax^2+bx+1\)
We are told that \(f_{max}(1)=a+b+1=3\), --> \(a+b=2\).
\(f_{max}\) is vertex of parabola and the \(x\) coordinate of vertex is \(-\frac{b}{2a}=1\) --> \(b=-2a\) --> \(a+b=a-2a=-a=2\) --> \(a=-2\) and \(b=4\).
You've got 3 points and 3 equations, so you could use a system of equations to solve. Coordinates: (1,3) Given, (0,1) Given, and (2,1) Implied. I got (2,1) by doing a really rough sketch and seeing that (2,1) is a mirror of (0,1) with (1,3) as the max.
3 systems in quadratic form: a(1)^2 + b(1) + c = 3 a(0)^2 + b(0) + c = 1 a(2)^2 + b(2) + c = 1
a + b + c = 3 c = 1 4a + 2b + c = 1
Solving this would yield a = -2, b = 4, c = 1. f(x) = -2x^2 + 4x + 1
I understand how you arrived at the f(x)=ax^2+bx+c, but you I don't fully understand how you arrived at 2x^2+4x+1. Could you please explain a little more?
I understand how you arrived at the f(x)=ax^2+bx+c, but you I don't fully understand how you arrived at 2x^2+4x+1. Could you please explain a little more?
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We have the function \(f(x)=ax^2+bx+c\).
Stem: the value of the function at x =0 is 1.
Substitute x by 0 --> \(f(0)=1=a*0^2+b*0+c\) --> \(c=1\), hence the function becomes: \(f(x)=ax^2+bx+1\).
Stem: A quadratic function f(x) attains a max of 3 at x =1.
The maximum (or minimum if a>0) of the quadratic function \(f(x)=ax^2+bx+c\) is the y coordinate (the value of f(x)) of the vertex of the given quadratic function, which is parabola.
The vertex of the parabola is located at the point \((-\frac{b}{2a},\) \(c-\frac{b^2}{4a})\).
Given: A quadratic function f(x) attains a max of 3 at x =1 --> coordinates of the vertex are (1, 3) --> \(-\frac{b}{2a}=1\), \(c-\frac{b^2}{4a}=3\)
So we have: \(a+b=2\) and: \(-\frac{b}{2a}=1\)
From this \(a=-2\) and \(b=4\). so the function is: \(f(x)=-2x^2+4x+1\)
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07 Jul 2014, 01:40
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18 Aug 2015, 19:51
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Re: A quadratic function f(x) attains a max of 3 at x =1, the [#permalink]
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11 Sep 2015, 23:03
Bunuel wrote:
GMATBLACKBELT wrote:
xcusemeplz2009 wrote:
A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is
a)-159, b)-110, c)-180, d)-105, e)-119
Pls explain
Not sure this would be on the GMAT, seems you should use calculus for this.
ax^2+bx+c=0
f(1)=3=a+b+c f(0)=1=c
f'(x)= 2xa+b.
since the max is 3, f'(1)=0. Thus, 0=2a+b--> b=-2a --> 2=a-2a --> -2=a
f(10) = -2*100+10(4)+1 --> -159
This can be solved in the following way too:
We have \(f(x)=ax^2+bx+c\).
\(f(0)=c=1\) --> \(f(x)=ax^2+bx+1\)
We are told that \(f_{max}(1)=a+b+1=3\), --> \(a+b=2\).
\(f_{max}\) is vertex of parabola and the \(x\) coordinate of vertex is \(-\frac{b}{2a}=1\) --> \(b=-2a\) --> \(a+b=a-2a=-a=2\) --> \(a=-2\) and \(b=4\).
\(f(x)=-2x^2+4x+1\) --> \(f(10)=-200+40+1=-159\)
Answer: A.
how should i understand that 1 is the vertex of parabola. is it the reason that x=0 ?please clarify this point
Re: A quadratic function f(x) attains a max of 3 at x =1, the [#permalink]
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14 Sep 2016, 14:45
Hello from the GMAT Club BumpBot!
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48% (02:09) correct
