Math / Calculus

QuestionLet $f$ be the function defined by $f(x)=\frac{3 x^{20}}{4 e^{x}+8 x^{20}}$ for $x>0$ . Which of the following is a horizontal asymptote to the graph of $f$ ? (A) $y=0$
B $y=\frac{3}{8}$ (C) $y=\frac{3}{4}$ (D) There is no horizontal asymptote to the graph of $f$ .

Studdy Solution

STEP 1

What is this asking? We need to find the horizontal asymptote of a function, which means figuring out what value the function gets close to as $x$ becomes super large. Watch out! Don't just plug in a large number for $x$ – we need to analyze the behavior of the function as $x$ approaches infinity.

STEP 2

1. Analyze the function's behavior as $x$ approaches infinity.
2. Simplify the function.
3. Determine the horizontal asymptote.

STEP 3

Alright, so we've got this function $f(x) = \frac{3x^{20}}{4e^x + 8x^{20}}$ .
We're looking for a horizontal asymptote, which means we want to see what happens to $f(x)$ as $x$ gets really, really big!

STEP 4

Think about the terms in the function.
We have $3x^{20}$ in the numerator and $4e^x + 8x^{20}$ in the denominator.
As $x$ gets massive, both $x^{20}$ and $e^x$ will grow very large, but $e^x$ grows much faster than any polynomial, including $x^{20}$ .

STEP 5

Since $e^x$ grows much faster than $x^{20}$ , the $8x^{20}$ term in the denominator becomes insignificant compared to $4e^x$ as $x$ approaches infinity.
So, for very large $x$ , our function effectively behaves like $f(x) \approx \frac{3x^{20}}{4e^x}$ .

STEP 6

Now, we can rewrite this as $f(x) \approx \frac{3}{4} \cdot \frac{x^{20}}{e^x}$ .
Remember how we said $e^x$ grows much faster than $x^{20}$ ?
That means as $x$ goes to infinity, the fraction $\frac{x^{20}}{e^x}$ goes to zero.

STEP 7

We've simplified our function to $f(x) \approx \frac{3}{4} \cdot \frac{x^{20}}{e^x}$ , and we know that $\frac{x^{20}}{e^x}$ approaches zero as $x$ gets huge.
So, what happens when we multiply $\frac{3}{4}$ by something that's getting closer and closer to zero?
The whole expression approaches zero!

STEP 8

Therefore, as $x$ approaches infinity, $f(x)$ approaches zero.
This means the horizontal asymptote is $y = 0$ .

STEP 9

The horizontal asymptote to the graph of $f$ is $y = 0$ , so the answer is (A).

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QuestionLet $f$ be the function defined by $f(x)=\frac{3 x^{20}}{4 e^{x}+8 x^{20}}$ for $x>0$ . Which of the following is a horizontal asymptote to the graph of $f$ ? (A) $y=0$
B $y=\frac{3}{8}$ (C) $y=\frac{3}{4}$ (D) There is no horizontal asymptote to the graph of $f$ .

Studdy Solution

STEP 1

What is this asking? We need to find the horizontal asymptote of a function, which means figuring out what value the function gets close to as $x$ becomes super large. Watch out! Don't just plug in a large number for $x$ – we need to analyze the behavior of the function as $x$ approaches infinity.

STEP 2

1. Analyze the function's behavior as $x$ approaches infinity.
2. Simplify the function.
3. Determine the horizontal asymptote.

STEP 3

Alright, so we've got this function $f(x) = \frac{3x^{20}}{4e^x + 8x^{20}}$ .
We're looking for a horizontal asymptote, which means we want to see what happens to $f(x)$ as $x$ gets really, really big!

STEP 4

Think about the terms in the function.
We have $3x^{20}$ in the numerator and $4e^x + 8x^{20}$ in the denominator.
As $x$ gets massive, both $x^{20}$ and $e^x$ will grow very large, but $e^x$ grows much faster than any polynomial, including $x^{20}$ .

STEP 5

Since $e^x$ grows much faster than $x^{20}$ , the $8x^{20}$ term in the denominator becomes insignificant compared to $4e^x$ as $x$ approaches infinity.
So, for very large $x$ , our function effectively behaves like $f(x) \approx \frac{3x^{20}}{4e^x}$ .

STEP 6

Now, we can rewrite this as $f(x) \approx \frac{3}{4} \cdot \frac{x^{20}}{e^x}$ .
Remember how we said $e^x$ grows much faster than $x^{20}$ ?
That means as $x$ goes to infinity, the fraction $\frac{x^{20}}{e^x}$ goes to zero.

STEP 7

We've simplified our function to $f(x) \approx \frac{3}{4} \cdot \frac{x^{20}}{e^x}$ , and we know that $\frac{x^{20}}{e^x}$ approaches zero as $x$ gets huge.
So, what happens when we multiply $\frac{3}{4}$ by something that's getting closer and closer to zero?
The whole expression approaches zero!

STEP 8

Therefore, as $x$ approaches infinity, $f(x)$ approaches zero.
This means the horizontal asymptote is $y = 0$ .

STEP 9

The horizontal asymptote to the graph of $f$ is $y = 0$ , so the answer is (A).

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Studdy Solution

STEP 1

STEP 2

1. Analyze the function's behavior as xxx approaches infinity.2. Simplify the function.3. Determine the horizontal asymptote.

STEP 3

Alright, so we've got this function f(x)=3x204ex+8x20f(x) = \frac{3x^{20}}{4e^x + 8x^{20}}f(x)=4ex+8x203x20​.We're looking for a horizontal asymptote, which means we want to see what happens to f(x)f(x)f(x) as xxx gets really, really big!

STEP 4

Think about the terms in the function.We have 3x203x^{20}3x20 in the numerator and 4ex+8x204e^x + 8x^{20}4ex+8x20 in the denominator.As xxx gets massive, both x20x^{20}x20 and exe^xex will grow very large, but exe^xex grows *much* faster than any polynomial, including x20x^{20}x20.

STEP 5

Since exe^xex grows much faster than x20x^{20}x20, the 8x208x^{20}8x20 term in the denominator becomes insignificant compared to 4ex4e^x4ex as xxx approaches infinity.So, for very large xxx, our function effectively behaves like f(x)≈3x204exf(x) \approx \frac{3x^{20}}{4e^x}f(x)≈4ex3x20​.

STEP 6

Now, we can rewrite this as f(x)≈34⋅x20exf(x) \approx \frac{3}{4} \cdot \frac{x^{20}}{e^x}f(x)≈43​⋅exx20​.Remember how we said exe^xex grows much faster than x20x^{20}x20?That means as xxx goes to infinity, the fraction x20ex\frac{x^{20}}{e^x}exx20​ goes to **zero**.

STEP 7

STEP 8

Therefore, as xxx approaches infinity, f(x)f(x)f(x) approaches **zero**.This means the horizontal asymptote is y=0y = 0y=0.

STEP 9

The horizontal asymptote to the graph of fff is y=0y = 0y=0, so the answer is (A).

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1. Analyze the function's behavior as $x$ approaches infinity.
2. Simplify the function.
3. Determine the horizontal asymptote.

Alright, so we've got this function $f(x) = \frac{3x^{20}}{4e^x + 8x^{20}}$ .
We're looking for a horizontal asymptote, which means we want to see what happens to $f(x)$ as $x$ gets really, really big!

Think about the terms in the function.
We have $3x^{20}$ in the numerator and $4e^x + 8x^{20}$ in the denominator.
As $x$ gets massive, both $x^{20}$ and $e^x$ will grow very large, but $e^x$ grows much faster than any polynomial, including $x^{20}$ .

Since $e^x$ grows much faster than $x^{20}$ , the $8x^{20}$ term in the denominator becomes insignificant compared to $4e^x$ as $x$ approaches infinity.
So, for very large $x$ , our function effectively behaves like $f(x) \approx \frac{3x^{20}}{4e^x}$ .

Now, we can rewrite this as $f(x) \approx \frac{3}{4} \cdot \frac{x^{20}}{e^x}$ .
Remember how we said $e^x$ grows much faster than $x^{20}$ ?
That means as $x$ goes to infinity, the fraction $\frac{x^{20}}{e^x}$ goes to zero.

Therefore, as $x$ approaches infinity, $f(x)$ approaches zero.
This means the horizontal asymptote is $y = 0$ .

The horizontal asymptote to the graph of $f$ is $y = 0$ , so the answer is (A).