Let, f(x) = x2 + bx + 7. If f'(5) = 2, then the value of b is
4
3
- 4
- 3
C.
- 4
Given, f(x) = x2 + bx + 7
On differentiating both sides w.r.t. x, we get
f'(x) = 2x + b
Also, f'(5) = 2
The function is not suitable to apply Rolle's theorem, since
f(x) is not continuous on [1, 5]
f(1) f(5)
f(x) is continuous only at x = 4
f(x) is not differentiable at x = 4