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The function f and g are defined by

$$ f(x) = (x-2)(4-x), 0 \le x \le 4 $$ $$ g(x) = |x-2|, 0 \lt x \le A $$

(i) Find the range of values of A for which the composite function $fg$ is defined.

(ii) If $ A = 5 $, find the range of values of x for which $ f(x) < g(x) $.

I have attempted the question already but I am unsure whether the answers are right or not.

Attempt (i):

In order for $fg$ to be defined, the range of g(x) must be within the domain of f(x).

Therefore, $ 0 \le g(x) \le 4 $.

$ 0 \le |x-2| \le 4 $

$ |x-2| \ge 0 $ or $ |x-2| \le 4 $

. . .

So, $ -2 \le x \le 6 $.

Since the domain of g(x) is $ 0 \lt x \le A $, so the domain fg must be $ 0 \lt x \le 6 $.

Am I right until this point? If so, $ A = 6 $. However, the question asked for range of values of A and not a specific value.

Attempt (ii):

$ f(x) < g(x) $

$ (x-2)(4-x) \lt |x-2| $

Therefore,

$ x-2 \gt (x-2)(4-x) $ or $ x-2 \lt -(x-2)(4-x) $.

. . .

So, $ x \lt 2, x \gt 3 $ or $ x \gt 5 $.

However if $ A = 5 $, then x cannot be more than 5.

Thus, $ x \lt 2 $ or $ 3 \lt x \le 5 $.

user573999
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1 Answers1

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Part (i) is almost done. You just have $0\lt x\le A\le6$, so $A\in(0,6]$.

For part (ii), you have two cases: $x>2$ or $x<2$. Note that if $x=2$, then $f(2)=g(2)=0$.

For $x>2$, $|x-2|=x-2$ and $(4-x)(x-2)\lt x-2$ or $4-x\lt 1$ or $3\lt x$. You also have $x\le 5$, so $x\in(3,5]$.

For $x<2$, $|x-2|=2-x$, $(4-x)(x-2)\lt 2-x$ or $x-4\lt1$ or $x\lt 5$. Remember that you are in the case $x\lt 2$, but also $x\gt 0$. So for this case you have $x\in (0,2)$. The final answer is the union of the two intervals.

Andrei
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