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The functions $f$ and $g$ are defined for $x \in \mathbb{R}$ as $f : x \mapsto 2x^2 - 3x + 2$ and $g : x \mapsto x^3 + x + 7$ - Leaving Cert Mathematics - Question 5 - 2012

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The-functions-$f$-and-$g$-are-defined-for-$x-\in-\mathbb{R}$-as-$f-:-x-\mapsto-2x^2---3x-+-2$---and--$g-:-x-\mapsto-x^3-+-x-+-7$-Leaving Cert Mathematics-Question 5-2012.png

The functions $f$ and $g$ are defined for $x \in \mathbb{R}$ as $f : x \mapsto 2x^2 - 3x + 2$ and $g : x \mapsto x^3 + x + 7$. (a) Find the co-ordinates of the t... show full transcript

Worked Solution & Example Answer:The functions $f$ and $g$ are defined for $x \in \mathbb{R}$ as $f : x \mapsto 2x^2 - 3x + 2$ and $g : x \mapsto x^3 + x + 7$ - Leaving Cert Mathematics - Question 5 - 2012

Step 1

Find the co-ordinates of the two points where the curves $y = f(x)$ and $y = g(x)$ intersect.

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Answer

To find the points of intersection, set the equations equal:

2x23x+2=x3+x+72x^2 - 3x + 2 = x^3 + x + 7

Rearranging gives:

x32x2+4x+5=0x^3 - 2x^2 + 4x + 5 = 0

To find the roots of this polynomial, we can apply the factor theorem or try possible rational roots. After testing x=1x = -1 and x=5x = 5, we arrive at:

  • For x=1x = -1:

ightarrow (-1, 7)$$

  • For x=5x = 5:

ightarrow (5, 37)$$

Thus, the co-ordinates where the curves intersect are the points (1,7)(-1, 7) and (5,37)(5, 37).

Step 2

Find the area of the region enclosed between the two curves.

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Answer

To find the area between the curves, we integrate the difference:

A=x1x2(g(x)f(x))dxA = \int_{{x_1}}^{{x_2}} (g(x) - f(x))dx where x1=1x_1 = -1 and x2=5x_2 = 5.

Calculating:

A=15(x3+x+7(2x23x+2))dxA = \int_{{-1}}^{{5}} (x^3 + x + 7 - (2x^2 - 3x + 2))dx

This simplifies to:

=15(x32x2+5x+5)dx= \int_{{-1}}^{{5}} (x^3 - 2x^2 + 5x + 5)dx

Evaluating the integral gives:

A=[x442x33+5x22+5x]15A = \left[ \frac{x^4}{4} - \frac{2x^3}{3} + \frac{5x^2}{2} + 5x \right]_{-1}^{5}

Calculating at x=5x = 5 and x=1x = -1, we obtain:

=(62542503+1252+25)(14+23525)= (\frac{625}{4} - \frac{250}{3} + \frac{125}{2} + 25) - (\frac{1}{4} + \frac{2}{3} - \frac{5}{2} - 5)

This results in an area of 36 square units, hence the final result is:

A=36A = 36.

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