SOLUTION: Given that the function f(x)= x/8 + 3/2x (i) Determine the turning points of the graph (ii) Determine the nature of the stationary points (iii) Hence sketch the graph.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Given that the function f(x)= x/8 + 3/2x (i) Determine the turning points of the graph (ii) Determine the nature of the stationary points (iii) Hence sketch the graph.      Log On


   

Question 1187848: Given that the function f(x)= x/8 + 3/2x
(i) Determine the turning points of the graph
(ii) Determine the nature of the stationary points
(iii) Hence sketch the graph.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+x%2F8+%2B+3%2F%282x%29+
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).
(i)
to find them, derivate
f'%28x%29+=+1%2F8+-+3%2F%282x%5E2%29
equal to zero and solve for+x
0=+1%2F8+-+3%2F%282x%5E2%29
3%2F%282x%5E2%29=+1%2F8.........cross multiply
24=+2x%5E2
x%5E2=12
x=sqrt%2812%29%29
x=sqrt%284%2A3%29
x=±2sqrt%283%29
solutions:
x=2sqrt%283%29 or x=-2sqrt%283%29

substitute in f%28x%29=+x%2F8+%2B+3%2F%282x%29+

f%28x%29=+%282sqrt%283%29%29%2F8+%2B+3%2F%282%2A2sqrt%283%29%29+
f%28x%29=+sqrt%283%29%2F2
or
f%28x%29=+%28-2sqrt%283%29%29%2F8+%2B+3%2F%28-2%2A2sqrt%283%29%29
f%28x%29=+-sqrt%283%29%2F2

turning points are at:
(2sqrt%283%29, sqrt%283%29%2F2)
(-2sqrt%283%29, -sqrt%283%29%2F2)
(ii) Determine the nature of the stationary points
(2sqrt%283%29, sqrt%283%29%2F2)->Minimum
(-2sqrt%283%29, -sqrt%283%29%2F2)->Maximum
(iii)