1I have done all that.Please care to inform me your final answer.I calculate it as Zero though book says one [251]
Let f(x)=\begin{Bmatrix} 1-\left|x \right| & \left| x\right|\leq 1 \\ 0& \left| x\right|\geq 1 \end{Bmatrix} and g(x)=\left| f(x+1)\right|+\left| f(x-1)\right|
Number of points where g(x) is discontinuous ?
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UP 0 DOWN 0 0 2
2 Answers
39First we define f(x) properly.
f(x) CASE 1 :
1 - |x|, |x| ≤ 1
or 1 - |x|, -1 ≤ x ≤ 1
or 1 + x, -1 ≤ x < 0
and 1 - x, 0 ≤ x ≤ 1
f(x) CASE 2 :
0, |x| ≥ 1
or 0 when x ≤ -1 U x ≥ 1
So our final function becomes
f(x) = \begin{Bmatrix} 0, & x < -1 \\ 1 + x, & -1 \leq x < 0 \\ 1 - x, & 0 \leq x \leq 1 \\ 0, & x > 1 \end{Bmatrix}
So accordingly we get
f(x+1)= \begin{Bmatrix} 0, & x < -2 \\ 2 + x & -2 \leq x < -1 \\ -x & -1 \leq x \leq 0 \\ 0 & x > 0 \end{Bmatrix}
f(x-1)= \begin{Bmatrix} 0, & x < 0 \\ x, & 0 \leq x < 1 \\ 2 - x, & 1 \leq x \leq 2 \\ 0, & x > 2 \end{Bmatrix}
Now how to apply the modulus to these piecewise functions..I forgot :D
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