The document discusses finding the maximum value of the function f(x) defined as the minimum of two quadratic expressions: -(x – 1)² + 2 and (x – 2)² + 1. The maximum value occurs at f(1) = 2 for x in the range (1, 2) and is lower outside this range. The conclusion is that the highest value of f(x) is 2, with specific intervals where each expression applies.

1. Functions

2. Functions
Find the maximum value of f(x); if f(x) is defined as the Min {-(x – 1)2
+ 2, (x – 2)2 + 1}.
(a) 1 (b) 2
(c) 0 (d) 3

3. Functions
First let us find the range where Min (-(x – 1)2 + 2, (x – 2)2 + 1) is –
(x – 1)2 + 2,
In other words, in which range is – (x – 1)2 + 2 < (x – 2)2 + 1.
–(x2 – 2x +1) + 2 < x2 – 4x + 4 + 1
0 < 2x2 – 6x + 4
x2 – 3x + 2 > 0
(x – 1) (x – 2) > 0
=> x > 2 or x < 1
Find the maximum value of f(x); if f(x) is defined as the Min {-(x – 1)2
+ 2, (x – 2)2 + 1}.

4. Functions
So, for x  (1, 2) , f(x) = (x – 2)2 + 1
And f(x) = –(x – 1)2 + 2 elsewhere.
Let us also compute f(1) and f(2)
f(1) = 2, f(2) = 1
For x  (-, 1), f(x) = –(x – 1)2 + 2
f(1) = 2
For x  (1, 2), f(x) = (x – 2)2 + 1
Find the maximum value of f(x); if f(x) is defined as the Min {-(x – 1)2
+ 2, (x – 2)2 + 1}.

5. Functions
f(2) = 1
For x  (2, ), f(x) = –(x – 1)2 + 2
For x < 1 and x > 2, f(x) is -(square) + 2 and so is less than 2.
When x lies between 1 and 2, the maximum value it can take is 2.
f(1) = 2 is the highest value f(x) can take.
Find the maximum value of f(x); if f(x) is defined as the Min {-(x – 1)2
+ 2, (x – 2)2 + 1}.

6. Functions
As a simple rule of thumb, the best way to approach this question is
to solve the two expressions. This gives us the meeting points of the
two curves. One of the two meeting points should be the maximum
value.
Answer choice (b)
Find the maximum value of f(x); if f(x) is defined as the Min {-(x – 1)2
+ 2, (x – 2)2 + 1}.

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