Question 1011338
 
Question:
If x can take any whole number value from 1 to 10,  find the smallest possible value of x2-8x+25.
 
Solution:
For {{{f(x)=x^2-8x+25}}}
Three possible ways.
  
1. By trial and error (x=1 to 10).  
We can form a table and find that at x=4, f(4)=16-32+25=9
  
2. By finding the vertex of the parabola, similar to completing the square.
f(x)=(x-4)^2+9, so at x=4, f(x)=9.
 
3. Using calculus.
f(x)=x^2-8x+25
f'(x)=2x-8=0 => x=4
f"(x)=2>0 => x=4 is a minimum