Answer:
The answer is "-2".
Step-by-step explanation:
Given value:
[tex]f(x) = 4(x-5)^2+2\\\\[/tex]
[tex]\ formula: \\\\(a-b)^2 = a^2+b^2-2ab\\\\[/tex]
[tex]\ f(x) = 4( x^2+5^2-2 \times 5 \times x) +2\\\\\ f(x) = 4(x^2+25-10x)+2\\\\\ f(x) = 4x^2+100-40x+2\\\\\ f(x) = 4x^2-40x+98\\\\[/tex]
taking derivative of the above function two times:
[tex]f (x)' = 8x-40\\\\f (x)'' = 8 \\\\[/tex]
let first derivative equal to 0.
[tex]8x-40 =0\\\\8x= 40\\\\x= \frac{40}{8}\\\\x= 5\\\\\ put \ the \ value \ of\ x \ in \ f(x)\ function :\\\\ f(x) = 4 (5)^2-40 (5)+98\\\\f(x)= 4 \times 25 - 40 \times 5 +98\\\\f(x) = 100-200+98\\\\f(x) = -100+98\\\\f(x) -2[/tex]
The greatest minimum value is -2.
Answer:
The parabola minimum is 3
The top quadratic minimum is 2 so
g(x) minimum is greater
Step-by-step explanation:
