Composite Function (Comparison Method) Example 3


Example 3:
Given f : xhx + k and f2 : x → 4x + 15.  
      (a)  Find the values of h and of k.
      (b)  Take h > 0, find the values of x for which f (x2) = 7x

Solution:
(a)
Step 1: find f2 (x)
Given f (x) = hx + k
f2 (x) = ff (x) = f (hx + k)
            = h (hx + k) + k
            = h2x + hk + k

Step 2: compare with given f2 (x)
f2 (x) = 4x + 15
h2x + hk + k = 4x + 15
h2 = 4
h = ± 2
When, h = 2
hk + k = 15
2k + k = 15
k = 5

When, h = –2
hk + k = 15
–2k + k = 15
k = –15

(b)
h > 0, h = 2, k = 5
Given f (x) = hx + k
f (x) = 2x + 5

f (x2) = 7x
2 (x2) + 5 = 7x
2x2 7x + 5 = 0
(2x 5)(x –1) = 0
2x 5 = 0      or     x –1= 0
x = 5/2                        x = 1