Answer:
Option B. minimum is correct for the first blank
Option C. 6 is correct for second blank.
Step-by-step explanation:
In order to find the maxima or minima of a function, we have to take the first derivative and then put it equal to zero to find the critical values.
Given function is:
[tex]f(x) = x^2-10x+31[/tex]
Taking first derivative
[tex]f'(x) = 2x-10[/tex]
Now the first derivative has to be put equal to zero to find the critical value
[tex]2x-10 = 0\\2x = 10\\x = \frac{10}{2} = 5[/tex]
The function has only one critical value which is 5.
Taking 2nd derivative
[tex]f''(x) =2[/tex]
[tex]f''(5) = 2[/tex]
As the value of 2nd derivative is positive for the critical value 5, this means that the function has a minimum value at x = 5
The value can be found out by putting x=5 in the function
[tex]f(5) = (5)^2-10(5)+31\\=25-50+31\\=6[/tex]
Hence,
The function y = x 2 - 10x + 31 has a minimum value of 6
Hence,
Option B. minimum is correct for the first blank
Option C. 6 is correct for second blank.
