If the function f:R→R defined by
f(x)={ax,x<2ax2−bx+3,x≥2 is differentiable, then the value of f′(−3)+f′(3) is equal to:
f(x)={ax;x<2ax2−bx+3;x≥2
Since f(x) is differentiable function
Then f(x) must be continuous.
Therefore, f(2+h)=f(2−h)=f(2)
⇒4a−2b+3=0
⇒2a−2b+3=0 ...(1)
Also, f′(2+h)=f′(2−h)
Thus 2a(2)−b=a
⇒3a=b ...(2)
Solving equation (1) & (2), we get
a=34 & b=94
Now f′(−3)=a
and f′(3)=2a(3)−b=6a−b
Therefore, f′(−3)+f′(3)=7a−b=214−94=124=3