I'm stuck on these two questions.
1)f(x) ≡ x^2 + 16/x, x ≠ 0.
a) Find f ′(x).
b) Find the coordinates of the stationary point of the curve y = f(x) and determine
f(x) = 4x^3 + ax^2 − 12x + b.
Given that a and b are constants and that when f(x) is divided by (x + 1) there is a
remainder of 15,
a find the value of (a + b).
Given also that when f(x) is divided by (x − 2) there is a remainder of 42,
b find the values of a and b,
c find the coordinates of the stationary points of the curve y = f(x).
(a point to anyone who doesnt know what " x^2 or x^3 " etc is - it means x to the power of 2 , 3 i,e x squared, cubed etc :P
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- Thread Starter
- 01-02-2006 21:59
- 01-02-2006 22:12
a) Remember (16/x)= 16x^-1
b) Stationary points occur when f ' (x) = 0. Find the value of x at this point and the corresponding value of y.
'Nature' refers to whether the stationary point is a min. max. or point of inflexion.
Find f ''(x) and insert the value for x:
If f '' (x) is:
negative - max. point
positive - min. point
zero - probably a point of inflexion
Offline20ReputationRep:TSR Group Staff
- TSR Group Staff
- 01-02-2006 22:22
"Given that a and b are constants and that when f(x) is divided by (x + 1) there is a remainder of 15"
This tells you that f(-1) = 15. You can do the division to find that out too.
f(x) = 4x3 + ax2 - 12x + b
15 = -4 + a + 12 + b
7 = a + b
If you apply the same logic with the next set of info (ie. f(2) = 42), you should end up with a pair of very simple simultaneous equations. Once you have a and b worked out, you can state f(x) in terms of x only. Then you can work out f'(x) = 0 for the stationary points.