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Exercise 9.1 · Q16

Q.Sketch the graph of ff, then identify the values of x0x_0 for which lim⁡x→x0f(x)\displaystyle\lim_{x\to x_0}f(x) exists.
[!FORMULA] f(x)={x2,x≤28−2x,2<x<44,x≥4f(x)=\begin{cases}x^2, & x\le2\\ 8-2x, & 2<x<4\\ 4, & x\ge4\end{cases}

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Step 1. Identify the pieces and their shapes. f(x)=x2f(x)=x^2 (a parabola) for x≤2x\le2; f(x)=8−2xf(x)=8-2x (a line, decreasing) for 2<x<42<x<4; f(x)=4f(x)=4 (a constant) for x≥4x\ge4. To sketch: draw the parabola up to and including x=2x=2, the descending line strictly between x=2x=2 and x=4x=4, and the flat line y=4y=4 from x=4x=4 onward.

Step 2. Check the first junction, x=2x=2. From the left (x≤2x\le2, using x2x^2): as x→2−x\to2^-, f(x)→22=4f(x)\to2^2=4. From the right (2<x<42<x<4, using 8−2x8-2x): as x→2+x\to2^+, f(x)→8−2(2)=4f(x)\to8-2(2)=4. Both sides agree at 44, so lim⁡x→2f(x)=4\displaystyle\lim_{x\to2}f(x)=4 exists (and it matches f(2)=4f(2)=4, so ff is even continuous there). …

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