Discover a wealth of knowledge and get your questions answered on IDNLearn.com. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.

Given the function:

[tex]\[ f(x)=\begin{cases}
\frac{1-\cos 6 x}{4 x^2}, & \text{if } x \neq 0 \\
A, & \text{if } x = 0
\end{cases} \][/tex]


Sagot :

Certainly! Let's derive the value of [tex]\( A \)[/tex] for the given function [tex]\( f(x) \)[/tex] so that it is continuous at [tex]\( x = 0 \)[/tex]. To ensure continuity, we need the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 0 to equal [tex]\( f(0) \)[/tex], which is [tex]\( A \)[/tex].

Given the function:
[tex]\[ f(x) = \begin{cases} \frac{1 - \cos 6x}{4x^2} & \text{if } x \neq 0 \\ A & \text{if } x = 0 \end{cases} \][/tex]

To find [tex]\( A \)[/tex], we evaluate the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos 6x}{4x^2} \][/tex]

To determine this limit, we can use L'Hôpital's Rule, which states that if we have a limit of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], we can differentiate the numerator and the denominator until the limit can be evaluated directly.

First, let's verify that the limit is indeed a [tex]\( \frac{0}{0} \)[/tex] form:
[tex]\[ \lim_{x \to 0} (1 - \cos 6x) = 1 - \cos 0 = 0 \][/tex]
[tex]\[ \lim_{x \to 0} 4x^2 = 4 \cdot 0^2 = 0 \][/tex]

Since it is a [tex]\( \frac{0}{0} \)[/tex] form, we can apply L'Hôpital's Rule:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos 6x}{4x^2} = \lim_{x \to 0} \frac{d}{dx} \left(1 - \cos 6x\right) \bigg/ \frac{d}{dx} (4x^2) \][/tex]

Now, we differentiate the numerator and the denominator:
[tex]\[ \frac{d}{dx} (1 - \cos 6x) = 6 \sin 6x \][/tex]
[tex]\[ \frac{d}{dx} (4x^2) = 8x \][/tex]

Applying L'Hôpital's Rule, we get:
[tex]\[ \lim_{x \to 0} \frac{6 \sin 6x}{8x} \][/tex]

We observe that this is still a [tex]\( \frac{0}{0} \)[/tex] form, so we apply L'Hôpital's Rule again:
[tex]\[ \lim_{x \to 0} \frac{6 \sin 6x}{8x} = \lim_{x \to 0} \frac{d}{dx} (6 \sin 6x) \bigg/ \frac{d}{dx} (8x) \][/tex]

Differentiating again:
[tex]\[ \frac{d}{dx} (6 \sin 6x) = 36 \cos 6x \][/tex]
[tex]\[ \frac{d}{dx} (8x) = 8 \][/tex]

Thus, the limit becomes:
[tex]\[ \lim_{x \to 0} \frac{36 \cos 6x}{8} \][/tex]

Evaluating the limit as [tex]\( x \to 0 \)[/tex]:
[tex]\[ \lim_{x \to 0} \frac{36 \cos 6x}{8} = \frac{36 \cdot \cos 0}{8} = \frac{36 \cdot 1}{8} = \frac{36}{8} = \frac{9}{2} \][/tex]

Therefore, for the function [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 0 \)[/tex], we need:
[tex]\[ A = \frac{9}{2} \][/tex]

So, the value of [tex]\( A \)[/tex] is [tex]\( \frac{9}{2} \)[/tex].