Question

Decide whether Rolle's Theorem can be applied to f(x) = x2 + 5 on the interval [0, 3]. If Rolle's Theorem can be applied, find all value(s) of c in the interval such that f ′(c) = 0.

Rolle's Theorem can be applied; c = negative one half
Rolle's Theorem can be applied; c = 0, 3
Rolle's Theorem cannot be applied because f ′(c) ≠ 0
Rolle's Theorem cannot be applied because f(0) ≠ f(3)

Answer 1

f(x)= x² + 5, is just a parabola shfited upwards by 5 units, so, is a smooth graph and no abrupt edges, so from 0 to 3, is indeed differentiable and continuous.  So Rolle's theorem applies, let's check for "c" by simply setting its variable to 0, bear in mind that, looking for "c" in this context, is really just looking for a critical point, since we're just looking where f'(c) = 0, and is a horizontal tangent line.

$\bf f(x)=x^2+5\implies \cfrac{df}{dx}=2x\implies 0=2x\implies \boxed{0=x}\leftarrow c$

Answer 2

The correct answer is option B which is Rolle's Theorem can be applied; c = 0, 3

What is Rolle's Theorem?

In calculus, Rolle's theorem or Rolle's lemma basically says that any real-valued differentiable function must have at least one stationary point or a point where the first derivative is zero, in order to have identical values at two different points.

f(x)= x² + 5, is just a parabola shifted upwards by 5 units, so, is a smooth graph and no abrupt edges, so from 0 to 3, is indeed differentiable and continuous.  

So Rolle's theorem applies, let's check for "c" by simply setting its variable to 0, bear in mind that, looking for "c" in this context, is really just looking for a critical point, since we're just looking where f'(c) = 0, and is a horizontal tangent line.

Therefore the correct answer is option B which is Rolle's Theorem can be applied; c = 0, 3

To know more about Rolle's theorem follow

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