Answer:
The difference between the maximum value of g(x) and the minimum value of f(x) is:
5
Step-by-step explanation:
- The function f(x) is given by:
[tex]f(x)=x^2-2x+8[/tex]
which in vertex form is given by:
[tex]f(x)=(x-1)^2+7[/tex]
Now, we know that:
[tex](x-1)^2\geq 0\\\\(x-1)^2+7\geq 0+7\\\\(x-1)^2+7\geq 7\\\\i.e.\\\\f(x)\geq 7[/tex]
Hence, the minimum value of the function f(x) is: 7
- Also, the function g(x) is a parabola that opens down and passes through (0,3) , (3,12) and (5,8) .
Let the equation for g(x) be:
[tex]g(x)=ax^2+bx+c[/tex]
Now with the help of the passing through three points we get:
when it passes through (0,3)
[tex]3=a\times 0+b\times 0+c\\\\i.e.\\\\c=3[/tex]
when it pass through (3,12)
i.e.
[tex]9a+3b+c=12\\\\i.e.\\\\9a+3b+3=12\\\\9a+3b=12-3\\\\9a+3b=9\\\\3a+b=3---------(2)[/tex]
Also it pass through (5,8)
i.e.
[tex]25a+5b+c=8\\\\25a+5b+3=8\\\\25a+5b=8-3\\\\25a+5b=5\\\\5a+b=1-------(2)[/tex]
On subtracting equation (2) from equation (1) we have:
[tex]a=-1[/tex]
and by putting the value of a in equation (1) we have:
[tex]b=6[/tex]
Hence, we get:
[tex]g(x)=-x^2+6x+3\\\\g(x)=-(x-3)^2+12[/tex]
Now, we know that:
[tex](x-3)^2\geq 0\\\\-(x-3)^2\leq 0\\\\-(x-3)^2+12\leq 12[/tex]
This means that the maximum value of g(x) is: 12
The difference between the maximum value of g(x) and the minimum value of f(x) is:
12-7=5