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Question
Verify Rolle's theorem for the following function on the indicated interval f (x) = x2 + 5x + 6 on the interval [−3, −2] ?
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Solution
Given function is \[f\left( x \right) = x^2 + 5x + 6\] .
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function,
\[f\left( x \right)\] is continuous and derivable on \[\left[ - 3, - 2 \right]\] .
Also,
\[f\left( - 3 \right) = \left( - 3 \right)^2 + 5\left( - 3 \right) + 6 = 9 - 15 + 6 = 0\]
\[f\left( - 2 \right) = \left( - 2 \right)^2 + 5\left( - 2 \right) + 6 = 4 - 10 + 6 = 0\]
\[ \therefore f\left( - 3 \right) = f\left( - 2 \right) = 0\]
Thus, all the conditions of the Rolle's theorem are satisfied.
Now, we have to show that there exists \[c \in \left[ - 3, - 2 \right]\] such that \[f'\left( c \right) = 0\] .
We have
\[f\left( x \right) = x^2 + 5x + 6\]
\[ \Rightarrow f'\left( x \right) = 2x + 5\]
\[ \therefore f'\left( x \right) = 0 \Rightarrow 2x + 5 = 0\]
\[ \Rightarrow x = \frac{- 5}{2}\]
Thus,
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