Answer :
A. [tex]f(x)=\frac{9x+7}{2x+4}\\f'(x)=\frac{(9)(2x+4)-(9x+7)(2)}{(2x+4)^2}\\f'(x)=\frac{(18x+36)-(18x+14)}{(2x+4)(2x+4)}\\f'(x)=\frac{22}{4x^2+16x+16}\\f'(x)=\frac{11}{2x^2+8x+8}\\\frac{11}{(2x+4)(x+2)}=0\\\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty[/tex] - There are no critical points because the graph is neither continuous nor smooth. There is a discontinuity at x = 2.
B. [tex]\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty[/tex] - The absolute maximum is f(lim⇒-2_-) = infinity. The absolute minimum is f(lim⇒-2_+) = -infinity. This applies to the interval [-10, 7].
C. [tex]f(x)=\frac{9x+7}{2x+4}\\f(0)=\frac{9(0)+7}{2(0)+4}\\f(0)=\frac{7}{4}\\f(0)=1.75\\f(5)=\frac{9(5)+7}{2(5)+4}\\f(5)=\frac{45+7}{10+4}\\f(5)=\frac{52}{14}\\f(5)=\frac{26}{7}\\f(5)=3.714[/tex] - The absolute maximum is f(5) = 26/7 or 3.714. The absolute mimimum is f(0) = 1.75. This applies to the interval [0, 5]. Proof: graph f(x) at [0, 5] on a graph or graphing calculator.
B. [tex]\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty[/tex] - The absolute maximum is f(lim⇒-2_-) = infinity. The absolute minimum is f(lim⇒-2_+) = -infinity. This applies to the interval [-10, 7].
C. [tex]f(x)=\frac{9x+7}{2x+4}\\f(0)=\frac{9(0)+7}{2(0)+4}\\f(0)=\frac{7}{4}\\f(0)=1.75\\f(5)=\frac{9(5)+7}{2(5)+4}\\f(5)=\frac{45+7}{10+4}\\f(5)=\frac{52}{14}\\f(5)=\frac{26}{7}\\f(5)=3.714[/tex] - The absolute maximum is f(5) = 26/7 or 3.714. The absolute mimimum is f(0) = 1.75. This applies to the interval [0, 5]. Proof: graph f(x) at [0, 5] on a graph or graphing calculator.
