Find the critical numbers, the intervals on which \( f(x) \) is increasing, the inte \( f(x)=\frac{x^{2}}{x-5} \) B. There is no solution. Select the correct choice below and, if necessary, fit in the answer box to A. The function is decreasing on \( (-\infty 0,0),(10, \infty) \), (Type your answer using interval nolation. Uso a comma to separa B. There is no solution. Which statement regarding the local extrema of the function is true? A. The function has a local minimum al \( x=0 \) and a local maximum at B. The function has a local minimum al \( x=10 \) and a local maximum C. The function has no local extrema.

We begin with \[ f(x)=\frac{x^2}{x-5}, \] with domain all real numbers except \( x=5 \). 1. **Finding the critical numbers**  a. Find \( f'(x) \) using the quotient rule. Recall that if \[ f(x)=\frac{g(x)}{h(x)}, \] then \[ f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2}. \]  b. Here, \( g(x)=x^2 \) and \( h(x)=x-5 \). Then, \[ g'(x)=2x \quad \text{and} \quad h'(x)=1. \]  c. Applying the quotient rule, we have \[ f'(x)=\frac{(2x)(x-5)-x^2(1)}{(x-5)^2} = \frac{2x(x-5)- x^2}{(x-5)^2}. \]  d. Expand the numerator: \[ 2x(x-5)=2x^2-10x, \] so \[ f'(x)=\frac{2x^2-10x-x^2}{(x-5)^2}=\frac{x^2-10x}{(x-5)^2}. \]  e. Factor the numerator: \[ x^2-10x=x(x-10). \] Thus, \[ f'(x)=\frac{x(x-10)}{(x-5)^2}. \]  f. Critical numbers occur when \( f'(x)=0 \) or is undefined (provided the point is in the domain). The numerator is zero when \[ x(x-10)=0, \] so \( x=0 \) or \( x=10 \). The derivative is undefined when the denominator is zero, i.e. when \[ (x-5)^2=0 \quad\Rightarrow\quad x=5. \] However, since \( f(x) \) is not defined at \( x=5 \) (vertical asymptote), we do not consider \( x=5 \) as a critical number.  g. Therefore, the critical numbers are \[ x=0 \quad \text{and} \quad x=10. \] 2. **Determining the intervals on which \( f(x) \) is increasing** Since \[ f'(x)=\frac{x(x-10)}{(x-5)^2}, \] and \( (x-5)^2 \) is always positive for \( x\neq5 \), the sign of \( f'(x) \) is determined by the numerator \( x(x-10) \).  a. Identify the zeros of the numerator: \[ x=0 \quad \text{and} \quad x=10. \]  b. Test intervals determined by these values (and taking care of the vertical asymptote at \( x=5 \)):   • For \( x<0 \) (for example, \( x=-1 \)):    \( (-1)\cdot((-1)-10) = (-1)\cdot(-11) = 11 > 0 \).    Thus, \( f'(x)>0 \) and \( f(x) \) is increasing on \( (-\infty,0) \).   • For \( 010 \) (for example, \( x=11 \)):    \( 11\cdot(11-10)=11\cdot(1)=11 > 0 \).    Thus, \( f(x) \) is increasing on \( (10,\infty) \).  c. Therefore, the function is increasing on \[ (-\infty,0) \quad \text{and} \quad (10,\infty). \] 3. **Local extrema**  a. At \( x=0 \):   - For \( x<0 \), \( f'(x)>0 \) (increasing).   - For \( 010 \), \( f'(x)>0 \) (increasing).   Thus, \( f(x) \) has a local minimum at \( x=10 \).   The function value is \[ f(10)=\frac{10^2}{10-5}=\frac{100}{5}=20. \] 4. **Answering the Multiple-Choice Statement** The statements provided are: - A. The function has a local minimum at \( x=0 \) and a local maximum at \(\dots\) - B. The function has a local minimum at \( x=10 \) and a local maximum at \( x=0 \). - C. The function has no local extrema. From our analysis, the correct statement is: \[ \textbf{B. The function has a local minimum at } x=10 \textbf{ and a local maximum at } x=0. \] 5. **Summary** - The critical numbers of \( f(x)=\frac{x^2}{x-5} \) are \( x=0 \) and \( x=10 \). - \( f(x) \) is increasing on \( (-\infty,0) \) and \( (10,\infty) \). - \( f(x) \) is decreasing on \( (0,5) \) and \( (5,10) \). - \( f \) has a local maximum at \( x=0 \) (with \( f(0)=0 \)) and a local minimum at \( x=10 \) (with \( f(10)=20 \)).