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MISCELLANEOUS EXERCISE-8 · Q65

Q.Discuss the continuity of the following function at the point or on the interval indicated against it. If discontinuous, identify the type of discontinuity and state whether it is removable; if removable, redefine the function so that it becomes continuous: f(x)=x2+2x+5f(x) = x^2+2x+5, for x≤3x \le 3, =x3−2x2−5= x^3-2x^2-5, for x>3x > 3.

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f(x)=x2+2x+5f(x)=x^2+2x+5 for x≤3x\le3, and f(x)=x3−2x2−5f(x)=x^3-2x^2-5 for x>3x>3. Here f(3)=9+6+5=20f(3)=9+6+5=20 (first piece).

Left-hand limit: lim⁡x→3−(x2+2x+5)=20\displaystyle\lim_{x\to3^-}(x^2+2x+5)=20 (matches f(3)f(3), trivially).

Right-hand limit: lim⁡x→3+(x3−2x2−5)=27−18−5=4\displaystyle\lim_{x\to3^+}(x^3-2x^2-5)=27-18-5=4. …

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