Math / Calculus

QuestionSketch the graph of the following function. Indicate where the function is increasing or decreasing, where any relative extrema occur, where asymptotes occur, where the graph is concave up or concave down, where any points of inflection occur, and where any intercepts occur. $f(x)=\frac{8 x-9}{x}$
On what interval(s) is fincreasing and on what interval(s) is decreasing? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function is increasing on $\square$ The function is never decreasing. (Simplify your answer. Type your answer in interval notation. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) B. The function is decreasing on $\square$ - The function is never increasing. (Simplify your answer. Type your answer in interval notation. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.) C. The function is increasing on $\square$ and decreasing on $\square$ 1. (Simplify your answers. Type your answers in interval notation. Type exact answers, using radicals as needed. Use a comma to separate answers as needed.) D. The function is never increasing or decreasing.

Studdy Solution

STEP 1

What is this asking? We need to sketch a graph of a function and find where it's increasing or decreasing. Watch out! Don't forget to consider what happens around $x = 0$ , since the function is undefined there!

STEP 2

1. Rewrite the function
2. Find the derivative
3. Analyze the derivative

STEP 3

Let's rewrite our function $f(x) = \frac{8x - 9}{x}$ in a more convenient form.
We can split the fraction into two parts: $f(x) = \frac{8x}{x} - \frac{9}{x}$

STEP 4

Now, since $\frac{x}{x} = 1$ (except when $x=0$ ), we can simplify this to: $f(x) = 8 - \frac{9}{x} = 8 - 9x^{-1}$ This form is much easier to work with when we take the derivative!

STEP 5

Now, let's find the derivative of $f(x)$ .
Remember, the derivative tells us the instantaneous rate of change of our function, which will help us figure out where it's increasing or decreasing.
Using the power rule, the derivative of $8$ is 0, and the derivative of $-9x^{-1}$ is $(-9) \cdot (-1)x^{-2} = 9x^{-2}$ .

STEP 6

So, our derivative is: $f'(x) = 9x^{-2} = \frac{9}{x^2}$

STEP 7

To find where $f(x)$ is increasing, we need to find where $f'(x)$ is positive.
Notice that $x^2$ is always positive when $x$ is not zero.
Since $9$ is also positive, our derivative $f'(x) = \frac{9}{x^2}$ is always positive (except at $x=0$ where it's undefined).

STEP 8

This means $f(x)$ is increasing everywhere except at $x=0$ .
In interval notation, this is $(-\infty, 0)$ and $(0, \infty)$ .

STEP 9

The function is increasing on $(-\infty, 0)$ and $(0, \infty)$ .
The function is never decreasing.
So the answer is A.

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Studdy Solution

STEP 1

What is this asking? We need to sketch a graph of a function and find where it's increasing or decreasing. Watch out! Don't forget to consider what happens around $x = 0$ , since the function is undefined there!

STEP 2

1. Rewrite the function
2. Find the derivative
3. Analyze the derivative

STEP 3

Let's rewrite our function $f(x) = \frac{8x - 9}{x}$ in a more convenient form.
We can split the fraction into two parts: $f(x) = \frac{8x}{x} - \frac{9}{x}$

STEP 4

Now, since $\frac{x}{x} = 1$ (except when $x=0$ ), we can simplify this to: $f(x) = 8 - \frac{9}{x} = 8 - 9x^{-1}$ This form is much easier to work with when we take the derivative!

STEP 5

STEP 6

So, our derivative is: $f'(x) = 9x^{-2} = \frac{9}{x^2}$

STEP 7

To find where $f(x)$ is increasing, we need to find where $f'(x)$ is positive.
Notice that $x^2$ is always positive when $x$ is not zero.
Since $9$ is also positive, our derivative $f'(x) = \frac{9}{x^2}$ is always positive (except at $x=0$ where it's undefined).

STEP 8

This means $f(x)$ is increasing everywhere except at $x=0$ .
In interval notation, this is $(-\infty, 0)$ and $(0, \infty)$ .

STEP 9

The function is increasing on $(-\infty, 0)$ and $(0, \infty)$ .
The function is never decreasing.
So the answer is A.

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Studdy Solution

STEP 1

What is this asking? We need to sketch a graph of a function and find where it's increasing or decreasing. Watch out! Don't forget to consider what happens around x=0x = 0x=0, since the function is undefined there!

STEP 2

1. Rewrite the function2. Find the derivative3. Analyze the derivative

STEP 3

Let's **rewrite** our function f(x)=8x−9xf(x) = \frac{8x - 9}{x}f(x)=x8x−9​ in a more convenient form.We can split the fraction into two parts: f(x)=8xx−9x f(x) = \frac{8x}{x} - \frac{9}{x} f(x)=x8x​−x9​

STEP 4

Now, since xx=1\frac{x}{x} = 1xx​=1 (except when x=0x=0x=0), we can simplify this to: f(x)=8−9x=8−9x−1 f(x) = 8 - \frac{9}{x} = 8 - 9x^{-1} f(x)=8−x9​=8−9x−1 This form is much easier to work with when we take the derivative!

STEP 5

STEP 6

So, our derivative is: f′(x)=9x−2=9x2 f'(x) = 9x^{-2} = \frac{9}{x^2} f′(x)=9x−2=x29​

STEP 7

STEP 8

This means f(x)f(x)f(x) is increasing everywhere *except* at x=0x=0x=0.In interval notation, this is (−∞,0)(-\infty, 0)(−∞,0) and (0,∞)(0, \infty)(0,∞).

STEP 9

The function is increasing on (−∞,0)(-\infty, 0)(−∞,0) and (0,∞)(0, \infty)(0,∞).The function is never decreasing.So the answer is **A**.

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What is this asking? We need to sketch a graph of a function and find where it's increasing or decreasing. Watch out! Don't forget to consider what happens around $x = 0$ , since the function is undefined there!

1. Rewrite the function
2. Find the derivative
3. Analyze the derivative

Let's rewrite our function $f(x) = \frac{8x - 9}{x}$ in a more convenient form.
We can split the fraction into two parts: $f(x) = \frac{8x}{x} - \frac{9}{x}$

Now, since $\frac{x}{x} = 1$ (except when $x=0$ ), we can simplify this to: $f(x) = 8 - \frac{9}{x} = 8 - 9x^{-1}$ This form is much easier to work with when we take the derivative!

So, our derivative is: $f'(x) = 9x^{-2} = \frac{9}{x^2}$

This means $f(x)$ is increasing everywhere except at $x=0$ .
In interval notation, this is $(-\infty, 0)$ and $(0, \infty)$ .

The function is increasing on $(-\infty, 0)$ and $(0, \infty)$ .
The function is never decreasing.
So the answer is A.