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Math  /  Calculus

Question(1 point) Please answer the following questions about the function f(x)=2x2x2−25f(x)=\frac{2 x^{2}}{x^{2}-25}
Instructions: - If you are asked for a function, enter a function. - If you are asked to find xx - or yy-values, enter either a number or a list of numbers separated by commas. If there are no solutions, enter None. - If you are asked to find an interval or union of intervals, use interval notation. Enter \{ \} if an interval is empty. - If you are asked to find a limit, enter either a number, I for ∞,−I\infty,-I for −∞-\infty, or DNED N E if the limit does not exist. (a) Calculate the first derivative of ff. Find the critical numbers of ff, where it is increasing and decreasing, and its local extrema. f′(x)=f^{\prime}(x)= □\square Critical numbers x=x= □\square Union of the intervals where f(x)f(x) is increasing □\square Union of the intervals where f(x)f(x) is decreasing □\square Local maxima x=x= □\square Local minima x=x= □\square (b) Find the following left- and right-hand limits at the vertical asymptote x=−5x=-5. lim⁡x→−5−2x2x2−25=\lim _{x \rightarrow-5^{-}} \frac{2 x^{2}}{x^{2}-25}= □\square lim⁡x→−5+2x2x2−25=?\lim _{x \rightarrow-5^{+}} \frac{2 x^{2}}{x^{2}-25}=? □\square Find the following left-and right-hand limits at the vertical asymptote x=5x=5. lim⁡x→5−2x2x2−25=?\lim _{x \rightarrow 5^{-}} \frac{2 x^{2}}{x^{2}-25}=? □\square lim⁡x→5+2x2x2−25=\lim _{x \rightarrow 5^{+}} \frac{2 x^{2}}{x^{2}-25}= □\square Find the following limits at infinity to determine any horizontal asymptotes. lim⁡x→−∞2x2x2−25=?∨lim⁡x→+∞2x2x2−25=?\lim _{x \rightarrow-\infty} \frac{2 x^{2}}{x^{2}-25}=? \quad \vee \quad \lim _{x \rightarrow+\infty} \frac{2 x^{2}}{x^{2}-25}=? □\square □\square (c) Calculate the second derivative of ff. Find where ff is concave up, concave down, and has inflection points. f′′(x)=□f^{\prime \prime}(x)=\square
Union of the intervals where f(x)f(x) is concave up □\square Union of the intervals where f(x)f(x) is concave down □\square □\square Inflection points x=x=

Studdy Solution

STEP 1

What is this asking? We need to analyze the function f(x)=2x2x2−25f(x) = \frac{2x^2}{x^2 - 25}, finding its derivative, critical points, increasing/decreasing intervals, local extrema, limits at vertical asymptotes, horizontal asymptotes, second derivative, concavity, and inflection points. Watch out! Don't forget to consider the points where the function is undefined when determining intervals of increase/decrease and concavity.

STEP 2

1. First Derivative and Critical Points
2. Increasing/Decreasing Intervals and Local Extrema
3. Limits at Vertical Asymptotes
4. Limits at Infinity and Horizontal Asymptotes
5. Second Derivative
6. Concavity and Inflection Points

STEP 3

**Calculate the first derivative** using the quotient rule: If f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, then f′(x)=g′(x)⋅h(x)−g(x)⋅h′(x)(h(x))2f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{(h(x))^2}. Here, g(x)=2x2g(x) = 2x^2 and h(x)=x2−25h(x) = x^2 - 25. So, g′(x)=4xg'(x) = 4x and h′(x)=2xh'(x) = 2x.

STEP 4

Therefore, f′(x)=(4x)(x2−25)−(2x2)(2x)(x2−25)2=4x3−100x−4x3(x2−25)2=−100x(x2−25)2.f'(x) = \frac{(4x)(x^2 - 25) - (2x^2)(2x)}{(x^2 - 25)^2} = \frac{4x^3 - 100x - 4x^3}{(x^2 - 25)^2} = \frac{-100x}{(x^2 - 25)^2}.

STEP 5

**Find the critical numbers** by setting f′(x)=0f'(x) = 0: −100x(x2−25)2=0\frac{-100x}{(x^2 - 25)^2} = 0. This fraction is zero when the numerator is zero, so −100x=0-100x = 0, which means x=0x = 0.

STEP 6

**Determine where** f(x)f(x) **is increasing and decreasing**. f′(x)f'(x) is undefined at x=−5x = -5 and x=5x = 5, and f′(x)=0f'(x) = 0 at x=0x = 0.
We examine the sign of f′(x)f'(x) in the intervals (−∞,−5)(-\infty, -5), (−5,0)(-5, 0), (0,5)(0, 5), and (5,∞)(5, \infty).

STEP 7

For x<−5x < -5, f′(x)>0f'(x) > 0, so f(x)f(x) is **increasing**. For −5<x<0-5 < x < 0, f′(x)>0f'(x) > 0, so f(x)f(x) is **increasing**. For 0<x<50 < x < 5, f′(x)<0f'(x) < 0, so f(x)f(x) is **decreasing**. For x>5x > 5, f′(x)<0f'(x) < 0, so f(x)f(x) is **decreasing**.

STEP 8

Since f′(x)f'(x) changes from positive to negative at x=0x = 0, there is a **local maximum** at x=0x = 0.

STEP 9

The **vertical asymptotes** are at x=−5x = -5 and x=5x = 5.

STEP 10

lim⁡x→−5−2x2x2−25=500−=−∞\lim_{x \rightarrow -5^-} \frac{2x^2}{x^2 - 25} = \frac{50}{0^-} = -\infty lim⁡x→−5+2x2x2−25=500+=∞\lim_{x \rightarrow -5^+} \frac{2x^2}{x^2 - 25} = \frac{50}{0^+} = \infty lim⁡x→5−2x2x2−25=500−=−∞\lim_{x \rightarrow 5^-} \frac{2x^2}{x^2 - 25} = \frac{50}{0^-} = -\inftylim⁡x→5+2x2x2−25=500+=∞\lim_{x \rightarrow 5^+} \frac{2x^2}{x^2 - 25} = \frac{50}{0^+} = \infty

STEP 11

lim⁡x→−∞2x2x2−25=lim⁡x→−∞21−25x2=21−0=2\lim_{x \rightarrow -\infty} \frac{2x^2}{x^2 - 25} = \lim_{x \rightarrow -\infty} \frac{2}{1 - \frac{25}{x^2}} = \frac{2}{1 - 0} = 2 lim⁡x→∞2x2x2−25=lim⁡x→∞21−25x2=21−0=2\lim_{x \rightarrow \infty} \frac{2x^2}{x^2 - 25} = \lim_{x \rightarrow \infty} \frac{2}{1 - \frac{25}{x^2}} = \frac{2}{1 - 0} = 2

STEP 12

The **horizontal asymptote** is y=2y = 2.

STEP 13

f′′(x)=ddx(−100x(x2−25)2)=100(3x2+25)(x2−25)3.f''(x) = \frac{d}{dx} \left( \frac{-100x}{(x^2 - 25)^2} \right) = \frac{100(3x^2 + 25)}{(x^2 - 25)^3}.

STEP 14

The **second derivative** is undefined at x=−5x = -5 and x=5x = 5.
The numerator is always positive.
The denominator is positive when x2>25x^2 > 25, which means x<−5x < -5 or x>5x > 5.
The denominator is negative when −5<x<5-5 < x < 5.

STEP 15

f(x)f(x) is **concave up** on (−∞,−5)(-\infty, -5) and (5,∞)(5, \infty). f(x)f(x) is **concave down** on (−5,5)(-5, 5).

STEP 16

There are **no inflection points** since f(x)f(x) is not defined at x=−5x = -5 and x=5x = 5.

STEP 17

(a) f′(x)=−100x(x2−25)2f'(x) = \frac{-100x}{(x^2 - 25)^2}; Critical numbers x=0x = 0; Increasing: (−∞,−5)∪(−5,0)(-\infty, -5) \cup (-5, 0); Decreasing: (0,5)∪(5,∞)(0, 5) \cup (5, \infty); Local maxima x=0x = 0; Local minima None (b) lim⁡x→−5−f(x)=−∞\lim_{x \rightarrow -5^-} f(x) = -\infty; lim⁡x→−5+f(x)=∞\lim_{x \rightarrow -5^+} f(x) = \infty; lim⁡x→5−f(x)=−∞\lim_{x \rightarrow 5^-} f(x) = -\infty; lim⁡x→5+f(x)=∞\lim_{x \rightarrow 5^+} f(x) = \infty; lim⁡x→−∞f(x)=2\lim_{x \rightarrow -\infty} f(x) = 2; lim⁡x→∞f(x)=2\lim_{x \rightarrow \infty} f(x) = 2 (c) f′′(x)=100(3x2+25)(x2−25)3f''(x) = \frac{100(3x^2 + 25)}{(x^2 - 25)^3}; Concave up: (−∞,−5)∪(5,∞)(-\infty, -5) \cup (5, \infty); Concave down: (−5,5)(-5, 5); Inflection points None

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