Question 1203773


a)  find f(-6), plug -6 into the function 

f(-6) = (-6)^2 + 5
f(-6) = 36 + 5
f(-6) = 41

b)  f(x) = 14,  find  x:

14 = x^2 + 5

Subtract 5 from both sides:

x^2 = 14 - 5
x^2 = 9

 take the square root of both sides:

x = ±sqrt(9)
x = ±3

So,  x = 3 and x = -3.

c) To find f(x + h), you plug x + h  for x:

f(x + h) = (x + h)^2 + 5

d)  We know f(x+h) and f(x) from above

{{{((f(x + h) - f(x)))/h }}}= {{{((x + h)^2 + 5 - (x^2 + 5))/h}}}

 simplify the  R H S

{{{(x + h)^2 + 5 - (x^2 + 5)/h}}}

 = {{{((x^2 + 2xh + h^2 + 5) - (x^2 + 5))/h}}}


 cancel x^2 and -x^2 we get

{{{(2xh + h^2)/h}}}


h(2x + h)/h



So,{{{ (f(x + h) - f(x))/h = 2x + h.}}}