# In order for the function f(x)=k(x-1)+x(k+3)+2 to be a constant function, what should be the value of k ?

Jul 9, 2016

$k = - \frac{3}{2}$

#### Explanation:

A constant function of x will have same value for any real value of x.

So f(0)=f(1)

We have

$f \left(x\right) = k \left(x - 1\right) + x \left(k + 3\right) + 2$

for x=0,

$f \left(0\right) = - k + 2$

for x=1

$f \left(1\right) = k \left(1 - 1\right) + 1 \left(k + 3\right) + 2 = k + 5$

Now $f \left(0\right) = f \left(1\right)$

$\implies - k + 2 = k + 5$

$2 k = 2 - 5 = - 3$

$\therefore k = - \frac{3}{2}$

Alternative

Differentiating f(x) w.r.t x

$f ' \left(x\right) = k + k + 3 = 2 k + 3$

f(x) being constat funtion f'(x)=0

$\therefore 2 k + 3 = 0 \implies k = - \frac{3}{2}$

Jul 9, 2016

k=0

#### Explanation:

$f ' = 2 k =$( constant )' = 0. So, k = 0.