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EXERCISE 8.1 · Q4

Q.Examine whether the function is continuous at the point indicated against it: f(x)=x3−2x+1f(x) = x^3 - 2x + 1, if x≤2x \le 2, =3x−2= 3x - 2, if x>2x > 2, at x=2x = 2.

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f(x)=x3−2x+1f(x)=x^3-2x+1 for x≤2x\le2, and f(x)=3x−2f(x)=3x-2 for x>2x>2. Since x=2x=2 belongs to the first piece, f(2)=23−2(2)+1=8−4+1=5f(2)=2^3-2(2)+1=8-4+1=5.

Left-hand limit: lim⁡x→2−f(x)=lim⁡x→2(x3−2x+1)=8−4+1=5\displaystyle\lim_{x\to2^-} f(x)=\lim_{x\to2}(x^3-2x+1)=8-4+1=5.

Right-hand limit: lim⁡x→2+f(x)=lim⁡x→2(3x−2)=6−2=4\displaystyle\lim_{x\to2^+} f(x)=\lim_{x\to2}(3x-2)=6-2=4. …

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