**GMATH**practice exercise (Quant Class 20)

The graphs of f(x) = x^3-x and g(x) = mx+n are represented in the figure given. If m and n are constants, what is the value of mn ?

(A) -4

(B) -2

(C) 2

(D) 4

(E) 6

Answer: [spoiler]______(D)__[/spoiler]

- fskilnik@GMATH
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00:00

**A**

**B**

**C**

**D**

**E**

The graphs of f(x) = x^3-x and g(x) = mx+n are represented in the figure given. If m and n are constants, what is the value of mn ?

(A) -4

(B) -2

(C) 2

(D) 4

(E) 6

Answer: [spoiler]______(D)__[/spoiler]

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)

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- fskilnik@GMATH
- GMAT Instructor
**Posts:**1449**Joined:**09 Oct 2010**Thanked**: 59 times**Followed by:**32 members

$$? = m \cdot n$$

$$f\left( x \right) = {x^3} - x = x\left( {{x^2} - 1} \right) = x\left( {x + 1} \right)\left( {x - 1} \right)$$

$$f\left( 2 \right) = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,A = \left( {2,6} \right)$$

$$f\left( x \right) = 0\,\,\,\, \Rightarrow \,\,\,\,x = - 1,0,\,{\rm{or}}\,\,1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,B = \left( { - 1,0} \right)$$

$$g\left( x \right) = mx + n$$

$${\rm{line}}\,\,g\,\,:\,\,\,\left\{ \matrix{

\,m = {\rm{slope}} = {{6 - 0} \over {2 - \left( { - 1} \right)}} = 2 \hfill \cr

\,B\, \in \,{\rm{graph}}\left( g \right)\,\,\,\, \Rightarrow \,\,\,\,0 = m \cdot \left( { - 1} \right) + n\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 2 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2 \cdot 2$$

The correct answer is (D).

We follow the notations and rationale taught in the

Regards,

Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)

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