Question

Find the point c in the curve f(x) = x^3 + x^2 + x + 1 in the interval [0, 1] where slope of a tangent to a curve is equals to the slope of a line joining (0,1)

(a) 0.64

(b) 0.54

(c) 0.44

(d) 0.34

I got this question in quiz.

This interesting question is from Lagrange’s Mean Value Theorem in section Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (b) 0.54

Best explanation: f(x) = x^3 + x^2 + x + 1

f(x) is continuous in given interval [0,1].

f’(x) = 3x^2+2x+1

Since, value of f’(x) is always finite in interval (0, 1) it is differentiable in interval (0, 1).

f(0) = 1

f(1) = 4

By mean value theorem,

f’(c) = 3c^2 + 2c + 1 = (4-1)/(1-0) = 3

⇒ c = 0.548,-1.215

Since c belongs to (0, 1) c = 0.54.