gradient of a curve

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January 9th 2009, 02:39 PM
ally79

gradient of a curve

Find the gradient of the curve y=2x^3 - 5x^2 + 46x + 87 at the point where it crosses the x axis.

I'm having real problems with this one as all the examples i've done before have given me a full set of co-ordinates.

I know that it will cross the x axis when y equals 0 so i have at least one co-ordinate. I differentiated it to 6x^2 - 10x + 46 I then thought about making 6x^2 - 10x = -46 and trying to get a value of x from that but i'm getting nowhere.

Any help on this is apprecciated
January 9th 2009, 03:08 PM
DeMath

$\left. {2{x^3} - 5{x^2} + 46x + 87 = 0{\text{ }}} \right| \times 4 \Leftrightarrow$

$\Leftrightarrow 8{x^3} - 20{x^2} + 184x + 348 = 0 \Leftrightarrow$

$\Leftrightarrow 8{x^3} + 27 - 20{x^2} + 45 + 184x + 276 = 0 \Leftrightarrow$

$\Leftrightarrow 8{x^3} + 27 - 5\left( {4{x^2} - 9} \right) + 92\left( {2x + 3} \right) = 0$

$\Leftrightarrow \left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right) - 5\left( {2x - 3} \right)\left( {2x + 3} \right) + 92\left( {2x + 3} \right) = 0.$
January 9th 2009, 03:13 PM
ally79

no idea what you've done or how it helps me,

sorry
January 9th 2009, 03:26 PM
DeMath

Quote:

Originally Posted by ally79

no idea what you've done or how it helps me,

sorry

$\left. {2{x^3} - 5{x^2} + 46x + 87 = 0{\text{ }}} \right| \times 4 \Leftrightarrow$

$\Leftrightarrow 8{x^3} - 20{x^2} + 184x + 348 = 0 \Leftrightarrow$

$\Leftrightarrow 8{x^3} + 27 - 20{x^2} + 45 + 184x + 276 = 0 \Leftrightarrow$

$\Leftrightarrow 8{x^3} + 27 - 5\left( {4{x^2} - 9} \right) + 92\left( {2x + 3} \right) = 0\Leftrightarrow$

$\Leftrightarrow\left( {2x + 3} \right)\left( {4{x^2} - 6x + 9} \right) - 5\left( {2x - 3} \right)\left( {2x + 3} \right) + 92\left( {2x + 3} \right) = 0 \Leftrightarrow$

$\Leftrightarrow \left( {2x + 3} \right)\left( {4{x^2} - 6x + 9 - 5\left( {2x - 3} \right) + 92} \right) = 0 \Leftrightarrow$

$\Leftrightarrow \left( {2x + 3} \right)\left( {4{x^2} - 16x + 116} \right) = 0 \Leftrightarrow$

$\Leftrightarrow \left[ \begin{gathered}2x + 3 = 0, \hfill \\{x^2} - 4x + 29 = 0; \hfill \\ \end{gathered} \right. \Leftrightarrow \left[ \begin{gathered}{x_1} = - \frac{3}{2}, \hfill \\{x_{2,3}} = 2 \pm i\sqrt 5 . \hfill \\ \end{gathered} \right.$

${x_1} = - \frac{3}{2}$ So $\left( { - \frac{3}{2};y\left( { - \frac{3}{2}} \right)=0} \right)$ it's the point where ${y\left( x \right) = 2{x^3} - 5{x^2} + 46x + 87}$ crosses the x axis.

Or I do not understand you?
January 10th 2009, 03:24 PM
DeMath

ally79, I wrote this post, if you have not yet realized that you need to do.

First, remember the definition of a gradient $\nabla F$ of the function $F\left( {{x_1},{x_2}, \ldots ,{x_n}} \right)$ :

The gradient $\nabla F$ of the function $F\left( {{x_1},{x_2}, \ldots ,{x_n}} \right)$ is a vector, whose projection on the axises of coordinates are partial derivatives of functions $F\left( {{x_1},{x_2}, \ldots ,{x_n}} \right)$ , i.e.

$\nabla F = \left( {\frac{{\partial F}}{{\partial {x_1}}},\frac{{\partial F}}{{\partial {x_2}}}, \ldots ,\frac{{\partial F}}{{\partial {x_n}}}} \right).$

You have to find $\left|{\nabla y(x)}\right|$ of $y=2{x^3} - 5{x^2} + 46x + 87$ at the point where it crosses the x axis, i.e. you need to find the absolute value of the derivative function $y=2{x^3} - 5{x^2} + 46x + 87$ at the point where it crosses the x axis.

Finally, you need to find this value: $\left| {\nabla y(x)} \right| = \left| {y'\left( { - \frac{3}{2}} \right)} \right|.$
January 10th 2009, 03:51 PM
skeeter

Quote:

Find the gradient of the curve y=2x^3 - 5x^2 + 46x + 87 at the point where it crosses the x axis.

using the rational root theorem and the intermediate value theorem, you can determine that there exists one real root at $x = -\frac{3}{2}$ .

(why teachers still make students do this is beyond me)

now ... evaluate $\frac{dy}{dx}$ at $x = -\frac{3}{2}$