Given:
[tex]f\mleft(x\mright)=5x^2-7x+6[/tex]
To find:
a) The coordinates of the turning point on the graph of y = f(x).
b) Show that the turning point is a minimum point.
Explanation:
a) Using the first derivative test,
[tex]\begin{gathered} f^{\prime}(x)=0 \\ 10x-7=0 \\ x=\frac{7}{10} \end{gathered}[/tex]
Substituting in the given function we get,
[tex]\begin{gathered} f(\frac{7}{10})=5(\frac{7}{10})^2-7(\frac{7}{10})+6 \\ =\frac{71}{20} \end{gathered}[/tex]
Therefore, the turning point is,
[tex](\frac{7}{10},\frac{71}{20})[/tex]
b) Differentiating with respect to x again,
[tex]f^{\prime}^{\prime}(x)=10>0[/tex]
Therefore, the function has a minimum value at the turning point.
Hence, the turning point is a minimum point.
