Consider the function for x-2-3- fx- -6 -x-1 x 3 -2x 2 -5x-6 x-1 -x 1- The value of c obtained by applying Rolle-s theorem for which f-c-0 is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Summation by Sigma Method

Consider the ...

Question

Consider the function for $x \in [- 2, 3]$
$f (x) = ⎧ ⎨ ⎩ \begin{matrix} - 6; x = 1 \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}; x \neq 1 \end{matrix}$ . The value of c obtained by applying Rolle's theorem for which $f^{'} (c) = 0$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1/2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

'c' does not exist

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

x = [- 2, 3]

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{3} - 2 x^{2} - 5 x + 6}{x - 1}, & i f x \neq 1 - 6 & i f x = 1 \end{matrix}

f (x) = {\begin{matrix} 2 x^{2} - 1, & if 1 \leq x \leq 4 151 - 30 x, & if 4 < x \leq 5 \end{matrix}

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} cos (\frac{1}{x}), x \neq 0 0, x = 0 \end{matrix}

[- 1, 1]

c

f (x) = {\begin{matrix} x^{2} + 1 & , when 0 \leq x \leq 1 3 - x & , when 1 < x \leq 2 \end{matrix}

f (x) = | x |, - 1 \leq x \leq 1

e^{x}

Join BYJU'S Learning Program