Find the values of a and b so that the function ƒ(x) = {(X^2+3x+a), whenx≤1; when x>1 is differentiable at each x є R

Question

Find the values of a and b so that the function

  \(f(x) = \begin{cases} (x^2+3x+a), & \quad \text{when}\,x≤1; \text{}\\ (bx+2), & \quad \text{when}\,x>1 \text{} \end{cases}\)  is differentiable at each x є R

Answer 1

It is given that f(x) is differentiable at each x є R 

For x ≤ 1, 

f(x) = x² + 3x + a i.e. a polynomial 

for x > 1, 

f(x) = bx + 2, which is also a polynomial

Since, a polynomial function is everywhere differentiable. Therefore, f(x) is differentiable for all x > 1 and for all x < 1. 

f(x) is continuous at x = 1

\(\lim\limits_{x \to 1^-}\) f(x) =  \(\lim\limits_{x \to 1^+}\) f(x) = f(1)

 \(\lim\limits_{x \to 1}\) (x² + 3x +  a) =  \(\lim\limits_{x \to 1}\) (bx+2) = 1 + 3 + a

1² + 3(1) + a = b (1) + 2 = 4 + a

4 + a = b + 2 

a – b + 2 = 0 …(1) 

As function is differentiable, therefore, 

LHD = RHD LHD at x = 1:

As, LHD = RHD 

Therefore, 

5 = b 

Putting b in (1), we get, 

a – b + 2 = 0 

a - 5 + 2 =0 

a = 3

Hence, 

a = 3 and b = 5