Explanations:
a) Given the equation of a curve expressed as:
[tex]y=x^3-5x^2+7x-2[/tex]
On differentiating the function, we will have:
[tex]\begin{gathered} \frac{dy}{dx}=3x^{3-1}-2(5)x^{2-1}+7 \\ \frac{dy}{dx}=3x^2-10x+7 \end{gathered}[/tex]
If the dy/dx = 0, then the x-ccordinates will be:
[tex]\begin{gathered} 3x^2-10x+7=0 \\ 3x^2-3x-7x+7=0 \\ 3x(x-1)-7(x-1)=0 \\ (3x-7)(x-1)=0 \\ x=\frac{7}{3}and\text{ 1} \end{gathered}[/tex]
Determine the turning points
If x = 7/3 then;
[tex]\begin{gathered} y=(\frac{7}{3})^3^{}-5(\frac{7}{3})^2+7(\frac{7}{3})-2 \\ y=\frac{343}{27}-\frac{245}{9}+\frac{49}{3}-\frac{2}{1} \\ y=\frac{343-735+441-54}{27} \\ y=-\frac{5}{27} \end{gathered}[/tex]
One of the turning points is (7/3, -5/27)
if x = 1 then;
[tex]\begin{gathered} y=x^3-5x^2+7x-2 \\ y=1^3-5(1)^2+7(1)-2 \\ y=1-5+7-2 \\ y=1 \end{gathered}[/tex]
The other turning point is at (1, 1)
b) To get the minimum and maximum point, we will find the second derivative of the function as shown:
[tex]\begin{gathered} f^{\doubleprime}(x)=6x-10 \\ \text{If x = 7/3} \\ f^{\doubleprime}(\frac{7}{3})=6(\frac{7}{3})-10 \\ f^{\doubleprime}(\frac{7}{3})=\frac{42}{3}-10 \\ f^{\doubleprime}(\frac{7}{3})=\frac{12}{3}=4>0 \\ \end{gathered}[/tex]
Since f''(x) > 0 at the point where x = 7/3, hence the turning point (7/3, -5/27) is at the minimum
At the point where x = 1,
[tex]\begin{gathered} f^{\prime}(1)=6(1)-10 \\ f^{\doubleprime}(1)=6-10 \\ f^{\doubleprime}(1)=-4<0 \end{gathered}[/tex]
Since f''(x) > 0 at the point where x = 1, hence the turning point (1, 1) is at the maximum point.
c) To sketch the turning points (1, 1), and (7/3, -5/27)
To get other points, at x = 2
[tex]\begin{gathered} y=2^3-5(2)^2+7(2)-2 \\ y=8-20+14-2 \\ y=0 \end{gathered}[/tex]
Another coordinate point will be at (2, 0)
