Math / Calculus

Question21. Consider the function $f(x)=\frac{x^{3}-4 x}{x^{3}+1}$ . (1) Find the domain, the $x$ - and $y$ -intercepts, (2) Find the asymptotes, (3) Find the intervals of increase and decrease. (Tip: The root for $8 x^{3}+3 x^{2}-4=0$ is $x \approx 0.68636$ .)

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x) = \frac{x^3 - 4x}{x^3 + 1}$ .
2. We need to find the domain, $x$ -intercepts, and $y$ -intercepts.
3. We need to find any asymptotes of the function.
4. We need to determine the intervals where the function is increasing or decreasing.

STEP 2

Find the domain of the function.
The domain of $f(x)$ is all real numbers except where the denominator is zero. Set the denominator equal to zero and solve for $x$ :
$x^3 + 1 = 0$

STEP 3

Solve the equation $x^3 + 1 = 0$ .
$x^3 = -1$
$x = -1$
Thus, the domain of $f(x)$ is all real numbers except $x = -1$ .

STEP 4

Find the $x$ -intercepts by setting the numerator equal to zero.
$x^3 - 4x = 0$

STEP 5

Factor the equation $x^3 - 4x = 0$ .
$x(x^2 - 4) = 0$

STEP 6

Solve for $x$ in the equation $x(x^2 - 4) = 0$ .
$x = 0 \quad \text{or} \quad x^2 - 4 = 0$
$x^2 = 4$
$x = \pm 2$
The $x$ -intercepts are $x = 0, 2, -2$ .

STEP 7

Find the $y$ -intercept by evaluating $f(0)$ .
$f(0) = \frac{0^3 - 4 \cdot 0}{0^3 + 1} = 0$
The $y$ -intercept is $(0, 0)$ .

STEP 8

Find the vertical asymptotes by setting the denominator equal to zero, which we already found in STEP_3.
The vertical asymptote is at $x = -1$ .

STEP 9

Find the horizontal asymptote by comparing the degrees of the numerator and the denominator.
Since the degrees of the numerator and denominator are both 3, the horizontal asymptote is the ratio of the leading coefficients.
$y = \frac{1}{1} = 1$

STEP 10

To find intervals of increase and decrease, find the derivative $f'(x)$ .
Use the quotient rule: if $f(x) = \frac{u(x)}{v(x)}$ , then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ .
Here, $u(x) = x^3 - 4x$ and $v(x) = x^3 + 1$ .

STEP 11

Find $u'(x)$ and $v'(x)$ .
$u'(x) = 3x^2 - 4$
$v'(x) = 3x^2$

STEP 12

Apply the quotient rule to find $f'(x)$ .
$f'(x) = \frac{(3x^2 - 4)(x^3 + 1) - (x^3 - 4x)(3x^2)}{(x^3 + 1)^2}$

STEP 13

Simplify the expression for $f'(x)$ .
$f'(x) = \frac{3x^5 + 3x^2 - 4x^3 - 4 - 3x^5 + 12x^3}{(x^3 + 1)^2}$
$f'(x) = \frac{3x^2 + 8x^3 - 4}{(x^3 + 1)^2}$

STEP 14

Find the critical points by setting $f'(x) = 0$ .
$3x^2 + 8x^3 - 4 = 0$

STEP 15

Use the tip provided: the root for $8x^3 + 3x^2 - 4 = 0$ is $x \approx 0.68636$ .

STEP 16

Determine the sign of $f'(x)$ on intervals determined by the critical points and vertical asymptote.
1. $(-\infty, -1)$
2. $(-1, 0.68636)$
3. $(0.68636, \infty)$

STEP 17

Test the sign of $f'(x)$ in each interval to determine where the function is increasing or decreasing.
- For $x < -1$ , choose $x = -2$ . - For $-1 < x < 0.68636$ , choose $x = 0$ . - For $x > 0.68636$ , choose $x = 1$ .

STEP 18

Evaluate $f'(x)$ at these test points to determine the sign.
- $f'(-2)$ is negative, so $f(x)$ is decreasing on $(-\infty, -1)$ . - $f'(0)$ is negative, so $f(x)$ is decreasing on $(-1, 0.68636)$ . - $f'(1)$ is positive, so $f(x)$ is increasing on $(0.68636, \infty)$ .

STEP 19

Summarize the intervals of increase and decrease.
- $f(x)$ is decreasing on $(-\infty, -1) \cup (-1, 0.68636)$ . - $f(x)$ is increasing on $(0.68636, \infty)$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

Find the domain of the function.The domain of f(x) f(x) f(x) is all real numbers except where the denominator is zero. Set the denominator equal to zero and solve for x x x:x3+1=0 x^3 + 1 = 0 x3+1=0

STEP 3

Solve the equation x3+1=0 x^3 + 1 = 0 x3+1=0.x3=−1 x^3 = -1 x3=−1x=−1 x = -1 x=−1Thus, the domain of f(x) f(x) f(x) is all real numbers except x=−1 x = -1 x=−1.

STEP 4

Find the x x x-intercepts by setting the numerator equal to zero.x3−4x=0 x^3 - 4x = 0 x3−4x=0

STEP 5

Factor the equation x3−4x=0 x^3 - 4x = 0 x3−4x=0.x(x2−4)=0 x(x^2 - 4) = 0 x(x2−4)=0

STEP 6

Solve for x x x in the equation x(x2−4)=0 x(x^2 - 4) = 0 x(x2−4)=0.x=0orx2−4=0 x = 0 \quad \text{or} \quad x^2 - 4 = 0 x=0orx2−4=0x2=4 x^2 = 4 x2=4x=±2 x = \pm 2 x=±2The x x x-intercepts are x=0,2,−2 x = 0, 2, -2 x=0,2,−2.

STEP 7

Find the y y y-intercept by evaluating f(0) f(0) f(0).f(0)=03−4⋅003+1=0 f(0) = \frac{0^3 - 4 \cdot 0}{0^3 + 1} = 0 f(0)=03+103−4⋅0​=0The y y y-intercept is (0,0) (0, 0) (0,0).

STEP 8

Find the vertical asymptotes by setting the denominator equal to zero, which we already found in STEP_3.The vertical asymptote is at x=−1 x = -1 x=−1.

STEP 9

Find the horizontal asymptote by comparing the degrees of the numerator and the denominator.Since the degrees of the numerator and denominator are both 3, the horizontal asymptote is the ratio of the leading coefficients.y=11=1 y = \frac{1}{1} = 1 y=11​=1

STEP 10

STEP 11

Find u′(x) u'(x) u′(x) and v′(x) v'(x) v′(x).u′(x)=3x2−4 u'(x) = 3x^2 - 4 u′(x)=3x2−4v′(x)=3x2 v'(x) = 3x^2 v′(x)=3x2

STEP 12

Apply the quotient rule to find f′(x) f'(x) f′(x).f′(x)=(3x2−4)(x3+1)−(x3−4x)(3x2)(x3+1)2 f'(x) = \frac{(3x^2 - 4)(x^3 + 1) - (x^3 - 4x)(3x^2)}{(x^3 + 1)^2} f′(x)=(x3+1)2(3x2−4)(x3+1)−(x3−4x)(3x2)​

STEP 13

STEP 14

Find the critical points by setting f′(x)=0 f'(x) = 0 f′(x)=0.3x2+8x3−4=0 3x^2 + 8x^3 - 4 = 0 3x2+8x3−4=0

STEP 15

Use the tip provided: the root for 8x3+3x2−4=0 8x^3 + 3x^2 - 4 = 0 8x3+3x2−4=0 is x≈0.68636 x \approx 0.68636 x≈0.68636.

STEP 16

Determine the sign of f′(x) f'(x) f′(x) on intervals determined by the critical points and vertical asymptote.1. (−∞,−1) (-\infty, -1) (−∞,−1)2. (−1,0.68636) (-1, 0.68636) (−1,0.68636)3. (0.68636,∞) (0.68636, \infty) (0.68636,∞)

STEP 17

STEP 18

STEP 19

Summarize the intervals of increase and decrease.- f(x) f(x) f(x) is decreasing on (−∞,−1)∪(−1,0.68636) (-\infty, -1) \cup (-1, 0.68636) (−∞,−1)∪(−1,0.68636). - f(x) f(x) f(x) is increasing on (0.68636,∞) (0.68636, \infty) (0.68636,∞).

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the domain of the function.
The domain of $f(x)$ is all real numbers except where the denominator is zero. Set the denominator equal to zero and solve for $x$ :
$x^3 + 1 = 0$

Solve the equation $x^3 + 1 = 0$ .
$x^3 = -1$
$x = -1$
Thus, the domain of $f(x)$ is all real numbers except $x = -1$ .

Find the $x$ -intercepts by setting the numerator equal to zero.
$x^3 - 4x = 0$

Factor the equation $x^3 - 4x = 0$ .
$x(x^2 - 4) = 0$

Solve for $x$ in the equation $x(x^2 - 4) = 0$ .
$x = 0 \quad \text{or} \quad x^2 - 4 = 0$
$x^2 = 4$
$x = \pm 2$
The $x$ -intercepts are $x = 0, 2, -2$ .

Find the $y$ -intercept by evaluating $f(0)$ .
$f(0) = \frac{0^3 - 4 \cdot 0}{0^3 + 1} = 0$
The $y$ -intercept is $(0, 0)$ .

Find the vertical asymptotes by setting the denominator equal to zero, which we already found in STEP_3.
The vertical asymptote is at $x = -1$ .

Find the horizontal asymptote by comparing the degrees of the numerator and the denominator.
Since the degrees of the numerator and denominator are both 3, the horizontal asymptote is the ratio of the leading coefficients.
$y = \frac{1}{1} = 1$

Find $u'(x)$ and $v'(x)$ .
$u'(x) = 3x^2 - 4$
$v'(x) = 3x^2$

Apply the quotient rule to find $f'(x)$ .
$f'(x) = \frac{(3x^2 - 4)(x^3 + 1) - (x^3 - 4x)(3x^2)}{(x^3 + 1)^2}$

Find the critical points by setting $f'(x) = 0$ .
$3x^2 + 8x^3 - 4 = 0$

Use the tip provided: the root for $8x^3 + 3x^2 - 4 = 0$ is $x \approx 0.68636$ .

Determine the sign of $f'(x)$ on intervals determined by the critical points and vertical asymptote.
1. $(-\infty, -1)$
2. $(-1, 0.68636)$
3. $(0.68636, \infty)$

Summarize the intervals of increase and decrease.
- $f(x)$ is decreasing on $(-\infty, -1) \cup (-1, 0.68636)$ . - $f(x)$ is increasing on $(0.68636, \infty)$ .