Question

If f and g are real functions defined by f(x) = x² + 7 and g(x) = 3x + 5, find the following:

${\displaystyle \frac{f\left(t\right)-f\left(5\right)}{t-5}}$, if t ≠ 5

Answer 1

f(x) = x² + 7

Substituting x = t in f(x), we get

f(t) = t² + 7  .......(i)

Considering the same function,

f(x) = x² + 7

Substituting x = 5 in f(x), we get

f(5) = (5)² + 7

= 25 + 7

= 32 .......(ii)

From equation (i) and (ii), we get

${\displaystyle \frac{f\left(t\right)-f\left(5\right)}{t-5}=\frac{t^2+7-32}{t-5}}$

= ${\displaystyle \frac{t^2-25}{t-5}}$

= ${\displaystyle \frac{t^2-5^2}{t-5}}$

But we know a² – b² = (a + b)(a – b)

So above equation becomes

${\displaystyle \frac{f\left(t\right)-f\left(5\right)}{t-5}=\frac{(t+5)(t-5)}{t-5}}$

Cancelling the like terms, we get

${\displaystyle \frac{f\left(t\right)-f\left(5\right)}{t-5}}$ = t + 5