ok can anybody help me?q.no.10 of M/J 2009
The function f is defined by f : x ? 2x2 ? 12x + 13 for 0 ? x ? A, where A is a constant.
(i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants.
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry.
(iii) When A has this value, find the range of f.
The function g is defined by g : x ? 2x2 ? 12x + 13 for x ? 4.
(iv) Explain why g has an inverse.
(v) Obtain an expression, in terms of x, for g?1(x).
thanks in advance
Nid to your rescue
I hope im not late
i) completing the square method 2(x-3)
2-5
ii) the line of symmetry is given by -b/2a or you can simply derivate and equate to 0
using the first method
-(-12)/ 2X2= 3
x=3 (line of symmetry passes through x=3)
second method
dy/dx= 4x-12
line of symmetry will pass through the maxima or minima as it's a quadratic
4x-12=0
x=3
u can use any method
iii) Just substitute 3 in the eqn
u get y= -5
now to decide whether this is max or min
sub 3 in d2y/dx2 = 4x
4X3=12 >0 hence minima
that mean all values of y will be greater than -5
y>=-5 is the range
iv) g has an inverse because it is a 1-1 function. It has a given set of domain.
v) y=2x
2-12x-13
We wrote this is the form (a-b)
2+c
2(x-3)
2-5
y=2(x-3)
2-5
make x the subject
y+5=2(x-3)
2y+5
----= (x-3)
2 2
/2} = (x-3))
/2} +3)
in terms of x, so u swap the y with x
g
-1(x)=
/2}+ 3)
