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

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GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
The figure above shows the representations of the graphs of the functi  [#permalink]

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

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

_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: The figure above shows the representations of the graphs of the functi  [#permalink]

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

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

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

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

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

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

$$? = - {1 \over 4} + 4 = 3.75$$

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

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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Re: The figure above shows the representations of the graphs of the functi  [#permalink]

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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 $$g(f(0)=-0.25+4=3.75$$.
Re: The figure above shows the representations of the graphs of the functi   [#permalink] 05 Mar 2019, 15:32
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