Question
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x^2+ 5 x + 6$ on the interval $[-3, -2]$
✓
Answer
Here, $f(x) = x^2+ 5 x + 6$ 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,
$f(x) = x^2+ 5x + 6$
$f(-3) = (-3)^2+5(-3) + 6$
$= 9 - 15 + 6$
$f(-3) = 0 ....(i)$
$f(-2) = (-2)^2+ 5(-2) + 6$
$= 4 - 10 + 6$
$f(-2) = 20 ....(ii)$
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,
$f(x) = x^2 + 5x + 6$
$f'(x) = 2x + 5$
$\Rightarrow f'(c) = 0$
$2c + 5 = 0$
$\text{c}=\frac{-5}{2}\in(-3,-2)$
So, Rolle's theorem is verified.
f(x) is continuous is [-3, -2] and f(x) is differentiable is (-3, -2) since it is a polynomial function.
Now,
$f(x) = x^2+ 5x + 6$
$f(-3) = (-3)^2+5(-3) + 6$
$= 9 - 15 + 6$
$f(-3) = 0 ....(i)$
$f(-2) = (-2)^2+ 5(-2) + 6$
$= 4 - 10 + 6$
$f(-2) = 20 ....(ii)$
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,
$f(x) = x^2 + 5x + 6$
$f'(x) = 2x + 5$
$\Rightarrow f'(c) = 0$
$2c + 5 = 0$
$\text{c}=\frac{-5}{2}\in(-3,-2)$
So, Rolle's theorem is verified.
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