Question

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

Answer 1

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($\frac{-1}{8}$)+$\frac{1}{4}$-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.