Question

I dont get this.. Can anyone help on this... I appreciate it.

Answer 1

7.$\frac{f(4 + h) - f(4)}{h} = 8$
   $h(\frac{f(4 + h) - f(4)}{h}) = h(8) \\f(4 + h) - f(4) = 8h$
   $(h + 4)^{2} + (4)^{2} = 8h \\(h + 4)(h + 4) + 16 = 8h \\h(h + 4) + 4(h + 4) + 16 = 8h$
   $h(h) + h(4) + 4(h) + 4(4) + 16 = 8h \\h^{2} + 4h + 4h + 16 + 16 = 8h$
   $h^{2} + 8h + 32 = 8h \\h^{2} + 32 = 0$
   $h = \frac{-(0) \± \sqrt{(0)^{2} - 4(1)(32)}}{2(1)}$
   $h = \frac{0 \± \sqrt{(0)^{2} - 4(1)(32)}}{2}$
   $h = \frac{0 \± \sqrt{-128}}{2}$
   $h = \frac{0 \± 8i\sqrt{2}}{2}$
   $h =\frac{8i\sqrt{2}}{2}$
   $h = 4i\sqrt{2}$

8.$f(x) = \frac{x^{2} + 4x - 32}{x - 4} \\f(x) = \frac{x^{2} + 8x - 4x - 32}{x - 4} \\f(x) = \frac{x(x) + x(8) - 4(x) - 4(8)}{x - 4} \\f(x) = \frac{x(x + 8) - 4(x + 8)}{x - 4} \\f(x) = \frac{(x - 4)(x + 8)}{x - 4} \\f(x) = x + 8$

$g(x) = x(2x - 5) - 2(x^{2} - 3x - 4) \\g(x) = x(2x) - x(5) - 2(x^{2}) + 2(3x) + 2(4) \\g(x) = 2x^{2} - 5x - 2x^{2} + 6x + 8 \\g(x) = 2x^{2} - 2x^{2} - 5x + 6x + 8 \\g(x) = x + 8$

The functions are equivalent.

9.$Slope: 3 \\Derivative: -8x + 11 \\3 = -8x = x + 11 \\-9x = 11 \\\frac{-9x}{-9} = \frac{11}{-9} \\x = -1\frac{2}{9} \\f(x) = -4x^{2} + 11x - 2 \\ f(x) = -4(-1\frac{2}{9}) + 11(-1\frac{2}{9}) - 2 \\f(x) = 4\frac{8}{9} - 13\frac{4}{9} - 2 \\f(x) = -8\frac{5}{9} - 2 \\f(x) = -10\frac{5}{9} \\y - y_{1} = m(x - x_{1}) \\y - (-11\frac{4}{9}) = 3(x - (-1\frac{2}{9}) \\y + 11\frac{4}{9} = 3(x + 1\frac{2}{9}) \\y + 10\frac{5}{9} = 3(x) + 3(1\frac{2}{9}) \\y + 10\frac{5}{9} = 3x + 3\frac{2}{3} \\y = 3x - 6\frac{8}{9}$