Question

verify mean value theorem for x^3 + x^2 - 6x for the interval [0,4]

Answer 1

↵Dear Student,

The given function is f (x) = x ³ +x²– 6x

f (x) is a polynomial function, so it continuous and differentiable everywhere

(i) f (x) being polynomial function, is continuous on [0, 4]

(ii) f (x) is differentiable on [0, 4]

Thus, all the conditions of Rolle’s Theorem are satisfied.

So, there must exist a real number c ε[0, 4] such that f '(c) = 0

Now, f '(c) = 0

⇒ 3c² +2c-6= 0

⇒ $$\begin{gathered} \mathrm{c}=\frac{-2\pm \sqrt{4+72}}{6} \\ =\frac{-2\pm 8.71}{6} \\ \mathrm{So}\mathrm{here}\mathrm{c}=1.11\left[\mathrm{neglecting}-\mathrm{ve}\mathrm{as}\mathrm{it}\mathrm{lies}\mathrm{beyond}\left[0,4\right]\right] \\ \mathrm{is}\mathrm{such}\mathrm{that}\mathrm{f}\text{'}\left(\mathrm{c}\right)=0 \end{gathered}$$

Hence, Rolle’s Theorem is verified.

Regards