Verify rolles theorem for the function f(x)=2x*3+x*2-4x-2

1.	Verify rolles theorem for the function f(x)=2x3+x2-4x-2
Answer» Answer: Step-by-step explanation: The GIVEN function f(x) = 2 $x^{3}$ + $x^{2}$ - 4x - 2 ...(i) ( when $\frac{-1}{2}$ ≤x≤ $\sqrt{2}$ ) ∴f( $\frac{-1}{2}$ )=2( $\frac{-1}{8}$ )+ $\frac{1}{<klux>4</klux>}$ -4( $\frac{-1}{2}$ )-2 = -1/4 +1/4+2-2 =0 ∴f( $\sqrt{2}$ ) = 2.( $(\sqrt{2}) ^{3}$ )+ $(\sqrt{2}) ^{2}$ -4. $\sqrt{2}$ -2 = 4. $\sqrt{2}$ +2-4. $\sqrt{2}$ -2=0 ∴f( $\frac{-1}{2}$ )=f( $\sqrt{2}$ ) Diff. (i) w.r. to 'x' both sides, we get f'(x) = 6 $x^{2}$ + 2X - 4 replace with 'C' then ∴ f'(c) = 6 $c^{2}$ + 2c - 4 By Rolle's Theorems, ∴ 6 $c^{2}$ + 2c - 4 =0 6 $c^{2}$ + 6c - 4c - 4 = 0 6c(c + 1) - 4(c + 1) = 0 (c + 1)(6c - 4) = 0 ∴c≠-1 ,c= $\frac{4}{6}$ = $\frac{2}{3}$ ∴c ∈[ $\frac{-1}{2}$ , $\sqrt{2}$ ] Hence, verified Rolle's theorem.