5 5.8 Use the graphs of \( f \) and \( g \) (not calculations) to solve \[ f(x) \cdot g(x) \geq 0 \] QUESTION 6 6.1 Evaluate the following limits: (a) \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \) (b) \( \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} \) (c) Let \( f(x)=\frac{2 x+12}{|x+6|} \) Determine: (i) \( \lim _{x \rightarrow 6^{+}} f(x) \) (ii) \( \lim _{x \rightarrow 6^{-}} f(x) \) (iii) Use (i) and (ii) to deduce \( \lim f(x) \) at \( x=6 \). 6.2 Let \[ f(x)=\left\{\begin{array}{ccc} \cos x & \text { if } & x<0 \\ 0 & \text { if } & x=0 \\ 1-x^{2} & \text { if } & x>0 \end{array}\right. \]

Let's break down the problem step by step. ### Question 5: Solve \( f(x) \cdot g(x) \geq 0 \) To solve the inequality \( f(x) \cdot g(x) \geq 0 \) using the graphs of \( f \) and \( g \), we need to identify the intervals where the product of the two functions is non-negative. 1. **Identify the zeros of \( f(x) \) and \( g(x) \)**: - Find the points where \( f(x) = 0 \) and \( g(x) = 0 \). These points will divide the x-axis into intervals. 2. **Determine the sign of \( f(x) \) and \( g(x) \)** in each interval: - For each interval created by the zeros, check the sign of \( f(x) \) and \( g(x) \). - The product \( f(x) \cdot g(x) \) will be non-negative if both functions are either positive or both are negative. 3. **Combine the intervals**: - List the intervals where \( f(x) \cdot g(x) \geq 0 \). ### Question 6: Evaluate the limits #### 6.1 Evaluate the following limits: (a) \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) \) To evaluate this limit, we can substitute \( x = 2 \): \[ \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) = (2^2 + 2)(3 \cdot 2^3 + 6) \] Calculating this step-by-step: 1. Calculate \( 2^2 + 2 = 4 + 2 = 6 \). 2. Calculate \( 3 \cdot 2^3 + 6 = 3 \cdot 8 + 6 = 24 + 6 = 30 \). 3. Multiply the results: \( 6 \cdot 30 = 180 \). Thus, \[ \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) = 180. \] (b) \( \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} \) Substituting \( u = -2 \): \[ \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} = \sqrt{(-2)^{2} + 3(-2) + 6} \] Calculating this step-by-step: 1. Calculate \( (-2)^{2} = 4 \). 2. Calculate \( 3(-2) = -6 \). 3. Combine: \( 4 - 6 + 6 = 4 \). 4. Take the square root: \( \sqrt{4} = 2 \). Thus, \[ \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} = 2. \] (c) Let \( f(x)=\frac{2 x+12}{|x+6|} \) (i) \( \lim _{x \rightarrow 6^{+}} f(x) \) For \( x \rightarrow 6^{+} \), \( |x+6| = x + 6 \): \[ \lim _{x \rightarrow 6^{+}} f(x) = \lim _{x \rightarrow 6^{+}} \frac{2x + 12}{x + 6} = \frac{2(6) + 12}{6 + 6} = \frac{12 + 12}{12} = \frac{24}{12} = 2. \] (ii) \( \lim _{x \rightarrow 6^{-}} f(x) \) For \( x \rightarrow 6^{-} \), \( |x+6| = x + 6 \): \[ \lim _{x \rightarrow 6^{-}} f(x) = \lim _{x \rightarrow 6^{-}} \frac{2x + 12}{x + 6} = \frac{2(6) + 12}{6 + 6} = \frac{12 + 12}{12} = \frac{24}{12} = 2. \] (iii) Since both one-sided limits are equal: \[ \lim_{x \rightarrow 6} f(x) = 2. \] #### 6.2 Evaluate the piecewise function For the function \[ f(x)=\left\{\begin{array}{ccc} \cos x & \text { if } & x<0 \\ 0 & \text { if } & x=0 \\ 1-x^{2} & \text { if } & x>0 \end{array}\right. \] To find the limit as \( x \) approaches 0: 1. From the left (\( x \rightarrow 0^{-} \)): \( f(x) = \cos x \rightarrow \cos(0) = 1 \). 2. From the right (\( x \rightarrow 0^{+} \)): \( f(x) = 1 - x^{2} \rightarrow 1 - 0^{2} = 1 \). Since both limits are equal: \[ \lim_{x \rightarrow 0} f(x) = 1. \] ### Summary of Results - \( f(x) \cdot g(x) \geq 0 \): Analyze using graphs. - \( \lim _{x \rightarrow 2}\left(x^{2}+x\right)\left(3 x^{3}+6\right) = 180 \) - \( \lim _{u \rightarrow-2} \sqrt{u^{2}+3 u+6} = 2 \) - \( \lim _{x \rightarrow 6^{+}} f(x) = 2 \) - \( \lim _{x \rightarrow 6^{-}} f(x) = 2 \) - \( \lim_{x \rightarrow 6} f(x) = 2 \) - \( \lim_{x \rightarrow 0} f(x) = 1 \)