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The figure above shows the representations of the graphs of the functi

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New post 05 Mar 2019, 13:57
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GMATH practice exercise (Quant Class 13)

Image

The figure above shows the representations of the graphs of the functions f and g, where f(x) = 2^(-0.5x) and g(x) = ax^2+b (a and b constants). What is the value of g(f(0)) ?

(A) 3
(B) 3.25
(C) 3.5
(D) 3.75
(E) 4

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Re: The figure above shows the representations of the graphs of the functi  [#permalink]

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New post 05 Mar 2019, 15:30
fskilnik wrote:
GMATH practice exercise (Quant Class 13)

Image

The figure above shows the representations of the graphs of the functions f and g, where f(x) = 2^(-0.5x) and g(x) = ax^2+b (a and b constants). What is the value of g(f(0)) ?

(A) 3
(B) 3.25
(C) 3.5
(D) 3.75
(E) 4


Image


\(? = g\left( {f\left( 0 \right)} \right)\)


\(\left[ {{\rm{blue}}} \right]\,\,\,\left( {0,f\left( 0 \right)} \right) \in {\rm{graph}}\left( f \right)\,\,\,\, \Rightarrow \,\,\,\,f\left( 0 \right) = {2^{ - {1 \over 2}\left( 0 \right)}} = 1\)


\(? = g\left( 1 \right) = a \cdot {1^2} + b = a + b\)


\(\left[ {{\rm{red}}} \right]\,\,\,\,\left\{ \matrix{
\left( { - 4,b} \right) \in {\rm{graph}}\left( f \right)\,\,\,\, \Rightarrow \,\,\,\,b = {2^{ - {1 \over 2}\left( { - 4} \right)}} = 4 \hfill \cr
\left( { - 4,0} \right) \in {\rm{graph}}\left( g \right)\,\,\,\,\,\mathop \Rightarrow \limits^{b = 4} \,\,\,\,\,0 = a \cdot {\left( { - 4} \right)^2} + 4\,\,\,\, \Rightarrow \,\,\,\,a = - {1 \over 4} \hfill \cr} \right.\)


\(? = - {1 \over 4} + 4 = 3.75\)


The correct answer is (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: The figure above shows the representations of the graphs of the functi  [#permalink]

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New post 05 Mar 2019, 15:32
Okay, this strikes me as not entirely GMAT-authentic, but I'll bite.

We want \(g(f(0))\), which means plugging 0 into the equation \(f(x)\) and then putting the result into the equation \(g(x)\). The first part is straightforward enough: \(f(0)=2^{-0.5(0)}=2^0=1\). But plugging that result into \(g(x)\) is less satisfying: \(g(1)=a(1^2)+b=a+b\).

Let's go back to the graph and take a look... We see that \((-4,0)\) is on the graph of \(g(x)\). (The parabola must be \(g(x)\), even if you don't know the graphs of these equations automatically, because the equation for \(f(x)\) can never equal zero.) Plugging this in to \(g(x)\) gives \(a(-4^2)+b=0\), which is \(16a+b=0\). From the dotted lines, it also appears that \(f(-4)=g(0)\). Plugging all of this in to the two functions gives \(2^{-0.5(-4)}=a(0^2)+b\), which means \(2^2=4=b\). Now \(16a+b=0\) becomes \(16a+4=0\), so \(a=-\frac{1}{4}\).

Finally, we can return to the earlier result that \(g(f(0)=g(1)=a+b\). With \(a=-0.25\) and \(b=4\), we get \(g(f(0)=-0.25+4=3.75\).
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Re: The figure above shows the representations of the graphs of the functi   [#permalink] 05 Mar 2019, 15:32
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