Answer to Question #102244 in Calculus for BIVEK

Question

For the function f, defined by f(x)=4x^3-4x^2-7x-2 there
exist a point C ∈ ]-1/2,2[ satisfying f′(c)=0.

Answer 1

According to Rolle's Theorem, suppose that a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b).Then if f(a)=f(b)=0, then there exists at least one point c in the open interval (a,b) for which f'(c)=0.

Here f(x) = 4x³-4x²-7x-2 and interval is [-1/2,2].Also f(x) is a polynomial function so it is continuous on [-1/2,2] and differentiable on the open interval (-1/2,2). Again f(-1/2)=0=f(2),so all the hypotheses of Rolle's Theorem are satisfied,so there exists c in (-1/2,2) such that f'(c)=0.

To find c :
Differentiating f(x)=4x³ -4x²-7x-2 with respect to x, we get f'(x)=12 x²-8 x-7.

Now, f'(c)=0  => 12 c² -8 c -7=0 => (c-7/6) (c-(-1/2)) = 0 => c= 7/6 and c=(-1/2)Since -1/2 does not belong to (-1/2 , 2) so c=7/6 in (-1/2 , 2).