Calculus Questions Online

Several parts to a calculus question?

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Intercepts
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x-intercept: Let y = 0 and solve for x:

0 = (x² + 1)/(x² – 2)
0 = x² + 1

Since we can’t have imaginary x-intercepts, there are no x-intercepts exist for y function.

y-intercept: Let x = 0 and solve for y:

y = (0² + 1)/(0² – 2)
y = -½

Hence, the y-intercept is -½.

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Asymptotes
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Vertical asymptote: Take the function from the denominator and set it to zero:

x² – 2 = 0
x = ±√2

Therefore, the two vertical asymtotes are x = √2 and x = -√2.

Horizontal: Set the limit for y function with x approaching to ∞ and -∞:

lim x → -∞ y = 1
lim x → ∞ y = 1

Since both limits are congruent, the horizontal asymptote is y = 1.

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Extrema
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The quotient rule is y’ = (g(x)f'(x) – f(x)g'(x))/g(x)², coming from y = f(x)/g(x).

By quotient rule, take a derivative of the function:

y’ = ((x² – 2)(2x) – (x² + 1)(2x))/(x² – 2)²
y’ = (2x³ – 4x – 2x³ – 2x)/(x² – 2)²
y’ = (-6x)/(x² – 2)²

Let y’ = 0 and solve for x:

0 = -6x/(x² – 2)²
0 = -6x
x = 0

Note that f'(x) < 0 at (-∞, 0) and f'(x) > 0 at (0, ∞). Therefore, the point at x = 0 is a relative max.

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Inflection
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Use quotient rule again! This gives:

y” = (24x²)/(x² – 2)³ – 6/(x² – 2)²

Let y” = 0 and solve for x again! Since this gives imaginary numbers, there are no points of inflection.

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(Finally) Domain
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{x|x ≠ ±√2}

I hope this helps!

MATH 150 – Calculus I with Review -Now Online