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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Verify Rolle's theorem for the following function on the indicated intervals

f(x) = x2 + 5 x + 6 on the interval [-3, -2]

✓

Answer

Here, f(x) = x2 + 5 x + 6 on [-3, -2]
f(x) is continuous is [-3, -2] and f(x) is differentiable is (-3, -2) since it is a polynomial function.
Now,
f(x) = x2 + 5x + 6
f(-3) = (-3)2 +5(-3) + 6
= 9 - 15 + 6
f(-3) = 0 ....(i)
f(-2) = (-2)2 + 5(-2) + 6
= 4 - 10 + 6
f(-2) = 20 ....(ii)
From equation (i) and (ii),
f(-3) = f(-2)
So, Rolle's theorem is applicable is [-3, -2], we have to show that
f'(c) = 0 as $\text{c}\in (-3,-2)$
Now,
f(x) = x2 + 5x + 6
f'(x) = 2x + 5
⇒ f'(c) = 0
2c + 5 = 0
$\text{c}=\frac{-5}{2}\in(-3,-2)$
So, Rolle's theorem is verified.

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