Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
### Problem 8:
Determine whether the function [tex]\( h(x)=x^3-2x \)[/tex] is even, odd, or neither.
Solution:
Recall that a function [tex]\( f \)[/tex] is said to be even if [tex]\( f(-x) = f(x) \)[/tex] and odd if [tex]\( f(-x) = -f(x) \)[/tex].
Let's first evaluate [tex]\( h(-x) \)[/tex].
[tex]\[ h(-x) = (-x)^3 - 2(-x) \][/tex]
[tex]\[ h(-x) = -x^3 + 2x \][/tex]
Now, compare [tex]\( h(-x) \)[/tex] with [tex]\( h(x) \)[/tex]:
[tex]\[ h(x) = x^3 - 2x \][/tex]
Check for evenness:
[tex]\[ h(-x) \neq h(x) \][/tex]
As [tex]\( -x^3 + 2x \neq x^3 - 2x \)[/tex], the function is not even.
Check for oddness:
[tex]\[ -h(x) = - (x^3 - 2x) \][/tex]
[tex]\[ -h(x) = -x^3 + 2x \][/tex]
[tex]\[ h(-x) = -h(x) \][/tex]
Since [tex]\( h(-x) = -h(x) \)[/tex], the function is odd.
Hence, we see that [tex]\( h(-x) = -h(x) \)[/tex] and we conclude that [tex]\( h(x) \)[/tex] is odd.
### Problem 9:
Find the limit of the function [tex]\( f(x)=\frac{e^{3x}-1}{x} \)[/tex] as [tex]\( x \rightarrow 0 \)[/tex].
Solution:
Observe that as [tex]\( x \rightarrow 0 \)[/tex],
[tex]\[ \frac{e^{3x} - 1}{x} \][/tex]
becomes an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
Therefore, we apply L'Hôpital's rule, which means differentiating the numerator and the denominator separately and then taking the limit of the new function.
First, differentiate the numerator and the denominator:
[tex]\[ \text{Numerator: } \frac{d}{dx} (e^{3x} - 1) = 3e^{3x} \][/tex]
[tex]\[ \text{Denominator: } \frac{d}{dx} x = 1 \][/tex]
Now, apply L'Hôpital's rule:
[tex]\[ \lim_{x \rightarrow 0} \frac{e^{3x} - 1}{x} = \lim_{x \rightarrow 0} \frac{3e^{3x}}{1} \][/tex]
Evaluate the limit:
[tex]\[ \lim_{x \rightarrow 0} 3e^{3x} = 3e^0 = 3 \][/tex]
Thus, the limit is:
[tex]\[ \boxed{3} \][/tex]
So, the limit of the function [tex]\( f(x) = \frac{e^{3x} - 1}{x} \)[/tex] as [tex]\( x \rightarrow 0 \)[/tex] is 3.
Determine whether the function [tex]\( h(x)=x^3-2x \)[/tex] is even, odd, or neither.
Solution:
Recall that a function [tex]\( f \)[/tex] is said to be even if [tex]\( f(-x) = f(x) \)[/tex] and odd if [tex]\( f(-x) = -f(x) \)[/tex].
Let's first evaluate [tex]\( h(-x) \)[/tex].
[tex]\[ h(-x) = (-x)^3 - 2(-x) \][/tex]
[tex]\[ h(-x) = -x^3 + 2x \][/tex]
Now, compare [tex]\( h(-x) \)[/tex] with [tex]\( h(x) \)[/tex]:
[tex]\[ h(x) = x^3 - 2x \][/tex]
Check for evenness:
[tex]\[ h(-x) \neq h(x) \][/tex]
As [tex]\( -x^3 + 2x \neq x^3 - 2x \)[/tex], the function is not even.
Check for oddness:
[tex]\[ -h(x) = - (x^3 - 2x) \][/tex]
[tex]\[ -h(x) = -x^3 + 2x \][/tex]
[tex]\[ h(-x) = -h(x) \][/tex]
Since [tex]\( h(-x) = -h(x) \)[/tex], the function is odd.
Hence, we see that [tex]\( h(-x) = -h(x) \)[/tex] and we conclude that [tex]\( h(x) \)[/tex] is odd.
### Problem 9:
Find the limit of the function [tex]\( f(x)=\frac{e^{3x}-1}{x} \)[/tex] as [tex]\( x \rightarrow 0 \)[/tex].
Solution:
Observe that as [tex]\( x \rightarrow 0 \)[/tex],
[tex]\[ \frac{e^{3x} - 1}{x} \][/tex]
becomes an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
Therefore, we apply L'Hôpital's rule, which means differentiating the numerator and the denominator separately and then taking the limit of the new function.
First, differentiate the numerator and the denominator:
[tex]\[ \text{Numerator: } \frac{d}{dx} (e^{3x} - 1) = 3e^{3x} \][/tex]
[tex]\[ \text{Denominator: } \frac{d}{dx} x = 1 \][/tex]
Now, apply L'Hôpital's rule:
[tex]\[ \lim_{x \rightarrow 0} \frac{e^{3x} - 1}{x} = \lim_{x \rightarrow 0} \frac{3e^{3x}}{1} \][/tex]
Evaluate the limit:
[tex]\[ \lim_{x \rightarrow 0} 3e^{3x} = 3e^0 = 3 \][/tex]
Thus, the limit is:
[tex]\[ \boxed{3} \][/tex]
So, the limit of the function [tex]\( f(x) = \frac{e^{3x} - 1}{x} \)[/tex] as [tex]\( x \rightarrow 0 \)[/tex] is 3.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.