Sunday

February 1, 2015

February 1, 2015

Posted by **Gabby** on Sunday, April 28, 2013 at 4:19pm.

a)find y and x intercepts

b) find the first derivative

c) find any critical values

d) find any local(relative) extrema

e) find second derivative

f) discuss the concavity

g) find any inflection points

Please show me how u got this!

- Calculus -
**Reiny**, Sunday, April 28, 2013 at 9:24pmI will assume you meant:

f(x) = (x^2 + 1)/(x^2 - 9)

= (x^2 + 1)/((x-3)(x+3))

a) for y intercepts , let x = 0

then y = 1/-9 = - 1/9

for x intercepts, let y = 0

(x^2 + 1)/(x^2 - 9) = 0

then x^2 + 1 = 0

x^2 = -1 , which has no real solutions

so there are no x-intercepts

b) using the quotient rule

dy/dx = ( (x^2-9)(2x) - (x^2+1)(2x))/(x^-9)^2

= -20x/(x^2 - 9)^2

c) --- critical values are found in many of the other parts.

d) for local extrema, dy/dx = 0

-20x/(x^2 - 9)^2 = 0

-20x = 0

x = 0 , then f(0) = -1/9

time to look at the graph using Wolfram, ( my other answers are confirmed here)

http://www.wolframalpha.com/input/?i=y+%3D+%28x%5E2+%2B+1%29%2F%28x%5E2+-+9%29

there is a local maximum at -1/9 , when x = 0

e) y '' = ((x^2 - 9)^2 (-20) + 20x(2)(x^2 - 9)(2x) )/(x^2 - 9)^4

= 60(x^2 + 3)/(x^2 - 9)^3

f) concave up or concave down depends on y''

if y'' is > 0 for a given x, then the curve is concave up at that value

if y'' is < 0 for .... concave down

While the numerator of y’’ is always positive,

the denominator of y’’ could be + or –

for -3 < x < 3 , the value of (x^2 – 9)^3 is negative, all all values yield a positive result.

So the curve is concave down for -3 < x < 3

For all other values it is concave up

Of course it is undefined for x = ± 3, giving us vertical asymptotes at

X = 3 and x = -3

g)

points of inflection occur when y'' = 0

60(x^2 + 3)/(x^2 - 9)^3 = 0

60(x^2 + 3) = 0

x^2 + 3 = 0

x^2 = -3

No solution, thus no point of inflection

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