Calculus help?

Question

Find all points of inflection for the function f(x) = ln(1+x^2)
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Find dy for y=sec3x
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Find the point on the graph of y= sqrt(x+1) closest to the point (3,0)

Answer 1

For the first one, f'(x) = 2x/(1+x²), and f''(x) = (2-2x²)/(1+x²)². Solve f''(x)= 0 and get x = 1 or x = -1. f(x) is concave down on (-∞,-1) and (1,∞) and concave up on (-1,1). And of course f(x) = 0.

For the second one, use the chain rule: dy/dx = sec(3x)tan(3x)•d(3x)/dx = 3sec(3x)tan(3x).

For the third one, you want to minimize the distance between (x, √(x+1)) and (3,0). So d = √[(x-3)² + (√(x+1)-0)²] = √(x² - 5x + 10).
Simpler to minimize d² = D = x² - 5x + 10.

So let D' = 2x - 5 = 0, x = 5/2.

Answer 2

To find the points of inflections, you need to find the 2nd derivative.

The first derivative is (2x) / (1 + x^2)

so by the quotient rule..the 2nd derivative is...

(2)(1+x^2) - (2x)(2x)
--------------------------
(1+x^2)^2

you can simiplfy but to find the zeros of this, all you need to do is find the zeros fo the numerator...

2 + 2x^2 - 4x^2 = 0
2- 2x^2 = 0
x^2 = 1
x = +1 and -1 ...plug it back into the original function to the get the coordinate (x,y) for the point of inflection

Answer 3

The middle one is 3sec3xtan3x . . . uses the chain rule.

The first one you would have to take the second derivative of the function and then find all the values of x where the second derivative equals 0.