Question

A curve has equation x3 - 5x2 + 7x - 2dyDifferentiate the function to obtaindxa) Find the x coordinates of the points where = 0 and hence the coordinates of thed.xturning points on the curve.dyb) With the aid of a table consider the sign of on either side of the turning points,dxdetermine whether the turning points are maximum or minimum points.c) Sketch the curve showing the turning points clearly and label any other points ofinterest.

Answer 1

SOLUTION

Write our the function given

To differentiate the function, we apply the differentiation rule

$y=x^n,\frac{dy}{dx}=nx^{n-1}$

Hence

$\begin{gathered} y=x^3-5x^2+7x-2 \\ \text{Then} \\ \frac{d}{dx}\mleft(x^3-5x^2+7x-2\mright) \end{gathered}$

Then Apply the sum and difference rule for derivative, we have

$\begin{gathered} =\frac{d}{dx}\mleft(x^3\mright)-\frac{d}{dx}\mleft(5x^2\mright)+\frac{d}{dx}\mleft(7x\mright)-\frac{d}{dx}\mleft(2\mright) \\ =3x^2-10x+7-0 \\ =3x^2-10x+7 \end{gathered}$

For dy/dx =0,we have

$3x^2-10x+7=0$

solve quadratic equation, we have

$\begin{gathered} 3x^2-3x-7x+7=0 \\ 3x(x-1)-7(x-1)=0 \\ (3x-7)(x-1)=0 \end{gathered}$

Equation each factor the zero, we have

$\begin{gathered} 3x-7=0,x-1=0 \\ 3x=7,x=1 \\ x=\frac{7}{3},1 \end{gathered}$

Hence

The x coordinates are

x= 7/3 and x=1

To obtain the coordinate of the turning point, we substitute into the equation given

$\begin{gathered} y=x^3-5x^2+7x-2 \\ \text{for x=7/3} \\ y=(\frac{7}{3})^3-5(\frac{7}{3})^2+7(\frac{7}{3})-2 \end{gathered}$

Then by simplification, we have

$y=-\frac{5}{27}$

Then, one of the turning point is

$(\frac{7}{3},-\frac{5}{27})$

Then, we substitute the other value of x,

$\begin{gathered} \text{for x=1} \\ y=x^3-5x^2+7x-2 \\ y=(1)^3-5(1)^2+7(1)-2 \\ y=1-5+7-2 \\ y=1 \\ \text{turning point =(1,1)} \end{gathered}$

Therefore the other turning point is (1,1)

The turning point are (7/3, -5/27) and (1,1)