| X = -1 | X = 0 | X = 2 | |
|---|---|---|---|
| f(x) | 3 | 6 | 0 |
| g(x) | 0 | 1 | -1 |
In each of the intervals $$(-1, 0)$$ and $$(0, 2)$$ the function $$(f-3g)''$$ never vanishes. Then the correct statement(s) is (are)
Let $$F(x) = f(x) - 3g(x)$$
$$\therefore$$ $$F( - 1) = 3$$, $$F(0) = 3$$ and $$F(2) = 3$$
So, $$F'(x)$$ will vanish at least twice in
$$( - 1,0) \cup (0,2)$$.
$$\because$$ $$F''(x) > 0$$ or $$ < 0$$, $$\forall x \in ( - 1,0) \cup (0,2)$$
Hence, $$f'(x) - 3g'(x) = 0$$ has exactly one solution in $$( - 1,0)$$ and one solution in $$(0,2)$$.
The function $$f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$$ has a local minimum or a local maximum at x =
Here, $$f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$$
$$ = \left\{ {\matrix{ { - 2x - (x + 2) + (x - 2),} & {when\,x \le - 2} \cr { - 2x + x + 2 + 3x + 2,} & {when\, - 2 < x \le - {2 \over 3}} \cr { - 4x,} & {when\, - {2 \over 3} < x \le 0} \cr {4x,} & {when\,0 < x \le 2} \cr {2x + 4,} & {when\,x > 2} \cr } } \right.$$
$$ = \left\{ {\matrix{ { - 2x - 4,} & {x \le - 2} \cr {2x + 4,} & { - 2 < x \le - 2/3} \cr { - 4x,} & { - {2 \over 3} < x \le 0} \cr {4x,} & {0 < x \le 2} \cr {2x + 4,} & {x > 2} \cr } } \right.$$
Graph for y = f(x) is shown as
