Function, Composite Function, Domain and Limit

1. \(f(x)=(x+5); g(x)=x^4 \Rightarrow (f \circ g)(x)= \\ (1) \ x^4+5 \\ (2) \ (x+5)^4 \\ (3) \ (x^4+5)^4 \\ (4) \ none \ of \ the \ above\)

sol. \((f\circ g)(x)= f(g(x))=f(x^4)=x^4+5\), choose (1)

2. \(f(x)={1 \over (x+2)}; g(x)=3x+4 \Rightarrow (g \circ f)(x)= g(f(x)) = \\ (1) \ 3x+6 \\ (2) \ 3x+10 \\ (3) \ 3\cdot {1 \over (x+2)} + 4 \\ (4) \ none \ of \ the \ above\)

sol. \((g \circ f)(x)= g(f(x)) = g(\frac{1}{x+2}) = 3\cdot \frac{1}{x+2} +4\), choose (3)

3. \(f(x)=x+2; \ g(x)=3x+4 \Rightarrow (g \circ f)^{-1}(x)= \\ (1) \ {x-6 \over 3} \\ (2) \ {x-10 \over 3 } \\ (3) \ 3\cdot {1 \over (x-2)} + 2 \\ (4) \ none \ of \ the \ above\)

sol. \((g \circ f)(x)= g(f(x)) = g(x+2) = 3\cdot (x+2) +4 = 3x+10\), 即 \(y=3x+10\), 利用 \(x,y\) 交換解 \(y\), 可得 \(x=3y+10 \Rightarrow 3y=x-10 \Rightarrow y= \frac{x-10}{3}\), 故 \((g \circ f)^{-1}(x)= \frac{x-10}{3}\), choose (2)

5. \(f(x)={(3x-5) \over |x^2-1|} \Rightarrow D_{f}= \\ (1) \ x \not= 1 \\ (2) \ x \not= 1 \ and \ x \not= -1 \\ (3) \ x \geq 1 \ or \ x \leq -1 \\ (4) \ none \ of \ the \ above\)

sol. \(x^2-1 \not= 0 \Rightarrow x \not=-1 \ and \ x \not= 1\), choose (2)

6. \(f(x)={{2x-3} \over \sqrt{x^2-x}} \Rightarrow D_{f}= \\ (1) \ x \not= 0 \ or \ 1 \\ (2) \ x > 1 \ and \ x < 0 \\ (3) \ x > 1 \ or \ x < 0 \\ (4) \ none \ of \ the \ above\)

sol. \(x^2-x \geq 0 \ and \ x^2-x \not= 0 \Rightarrow x^2-x >0\), 先解 \(x^2-x =0\), 得 \(x=0\) 或 \(x=1\)，接下來帶入值判斷 \(x<0\), \(0<x<1\), \(x>1\) 三個區間中，那些區間可以滿足 \(x^2-x>0\), 得 \(x<0\) 或 \(x>1\) 可滿足，故選擇 (3)

7. \(f(x)=\sqrt{x^2-81}\) 的定義域

sol. \(x^2-81 \geq 0\), 先解 \(x^2-81 = 0\), 得 \(x=-9\) 或 \(x=9\) ，接下來帶入值判斷 \(x<-9\), \(-9<x<9\), \(x>9\) 三個區間中，那些區間可以滿足 \(x^2-81 \geq 0\), 得 \(x<-9\) 或 \(x>9\) 可滿足，因等號成立時亦滿足，故選擇 (3)

8. \(f(x)=\sqrt{16-x^2}\) 的定義域

sol. \(x^2-81 \geq 0\), 先解 \(16-x^2 = 0\), 得 \(x=-4\) 或 \(x=4\) ，接下來帶入值判斷 \(x<-4\), \(-4<x<4\), \(x>4\) 三個區間中，那些區間可以滿足 \(16-x^2 \geq 0\), 得 \(-4<x<4\) 可滿足，因等號成立時亦滿足，故選擇 (2)

9. \[\lim_{x \to 3} \sqrt[3]{8-8x^2}= \sqrt[3]{8-72}= \sqrt[3]{-64} =-4\]

10. \[\lim_{x \to 4} {{x^2-2x-8} \over {4x^2-16x}}= \lim_{x \to 4} {{(x-4)(x+2)} \over {4x(x-4)}}= \lim_{x \to 4} \frac{x+2}{4x} = \frac{6}{16} = \frac{3}{8}\]

11. \[\lim_{x \to -2^{-}} {{5x^2+12x+3} \over {2x^2+9x+10}}= \lim_{x \to -2^{-}} {{5x^2+12x+3} \over {(x+2)(2x+5)}}= \frac{-1}{0^{-}} = \infty\] (\(0^{-}\) 代表很接近 \(0\) 的負數)

12. \[\lim_{x \to 0^{-}} {{3x^2-5|x|} \over {4x}}= \lim_{x \to 0^{-}} {{3x^2-5*(-x)} \over {4x}}=\lim_{x \to 0^{-}} {{3x^2+5x} \over {4x}}= \lim_{x \to 0^{-}} {{3x+5} \over {4}} = \frac{5}{4}\] (\(x \to 0^{-}\) 代表 \(x\) 為負值，故去掉絕對值符號時要加上負號)

13. \[\lim_{x \to 0^{-}} {{|4x|-5|x|} \over {3x}} = \lim_{x \to 0^{-}} {{-4x+5x} \over {3x}} = \lim_{x \to 0^{-}} {{x} \over {3x}}= \frac{1}{3}\] (\(x \to 0^{-}\) 代表 \(x\) 為負值，故去掉絕對值符號時要加上負號)

14. \[\lim_{x \to 4^{+}}{ {16-x^2} \over {x^2-8x+16}} = \lim_{x \to 4^{+}}{ {-(x-4)(x+4)} \over {(x-4)^2}} = \lim_{x \to 4^{+}}{ {-(x+4)} \over {x-4}} = \frac{-8}{0^{+}} = -\infty\] (\(0^{+}\) 代表很接近 \(0\) 的正數)

15. \[\lim_{x \to 0^{-}}{ {\sqrt{3x^4+5x^2}} \over {x}} = - \lim_{x \to 0^{-}} \sqrt{\frac{3x^4+5x^2}{x^2}} = - \lim_{x \to 0^{-}} \sqrt{\frac{3x^2+5}{1}} = -\sqrt{5}\] (\(x \to 0^{-}\) 代表 \(x\) 為負值，故 \(x\) 放入根號時，根號外面要加上負號)

16. \[\lim_{x \to 0^{-}}{ {\sqrt{2x^5+4x^2}} \over {4x}} = - \lim_{x \to 0^{-}} \sqrt{\frac{2x^5+4x^2}{16x^2}} = - \lim_{x \to 0^{-}} \sqrt{\frac{2x^3+4}{16}} = -\frac{1}{2}\] (\(x \to 0^{-}\) 代表 \(x\) 為負值，故 \(4x\) 放入根號時，根號外面要加上負號)