The particle’s position as a function of time is given as $x=5t^2-9t+5$. Find out the minimum value of position co-ordinate? 

Given function  $x=5t^2-9t+3$
1st derivative,   $\frac{dx}{dt}=10t-9=0\therefore t=\frac{9}{10}=0.9\sec$
2nd derivative test, $\frac{d^{2}x}{d{t}^2}=10>0$
As second derivative is positive, hence there exists a minima at  $t=0.9\sec$
Now, check for the limiting values
When $t=0;x=3m$ and as $t\to \infty ,x\to \infty$
It means the maximum position coordinate does not exist
For minimum position, putting t = 0.9 sec in the equation 
 $x=5{\left(0.9\right)}^2-9\left(0.9\right)+3=-1.05m$
Now let us plot the graph