Calculus and critical points?

Question

I need to find the critical points of f(x) = (x^2 + x + 1)/(x^2 + 1)......
Please show particularly clearly how you find the first and 2nd derivatives of the function......pretty please and thanks

Answer 1

Hey Darren, Colm here! I dont have a clue what the answer to your question is, but i sure wish i knew it today for the exam...doh!!!

Answer 2

Use the quotient rule to find the derivative of the fraction. Use the addition rule, the power rule, and the constant multiplier rule to find the derivatives of the numerator and denominator. Differentiate the result to get the second derivative. Finally, the critical points[1] are those where the (first) derivative is either zero or undefined.

Hope this helps.

Answer 3

f(x) = (x^2 + x + 1)(x^2 + 1) multiply together
f(x) = (x^4 +X^3 + x^2 +x^2 + x +1) collect like terms
f(x) = x^4 + x^3 +2x^2 + x +1
f'(x) = 4x^3 + 3x^2 +4x + 1 1st derivative
f"(x) = 12x^2 +6x + 4 2nd derivative

When f'(x) = 0 then these are the maximum and minimum points.

When f"(x) = 0 then these are the points of inflexion.
The 'Inflexion' point - when a curve is rising to its max. points the curve is becoming less steeper.After passing its max. points it comes more steeper as it falls to its minimum point. However its reaches a point in its fall to the minimum point where it is becoming less steeper. This point in known as the 'POINT OF INFLEXION'. In curves which trace the opposite way then the points of inflexion still hold - its just the trace is of the opposite pattern.

Answer 4

do the first derivative to get:
f'(x)= (2x +1) / (2x)......set the func = to zero and u get that the cp is : x=1/2 ...plug it into origial equation and u get the poit is ((1/2), (7/5)

next do the second derivative to get

f''(x)= 2/2 or 1.

Answer 5

y = (x^2 + x + 1)/(x^2 + 1) = ((x^2 + 1) + x)/(x^2 + 1) =1 + x/(x^2 + 1);
g(x) =xx+1, g’=2x;
y’ = 1’ + (x’g –g’x) /(gg) = 0+ (1*g –(2x)x)/ (gg) = (xx +1 –2xx)/(gg) = (1 -xx)/(xx+1)^2;
h(x) = 1-xx; h’ =-2x; y’ =h*g^(-2);
y’’ =h’*g^(-2) +h*{(-2)g^(-3) * g’} = (h’g –2hg’)/g^3 =
=(-2x(xx+1) –2(1-xx)*2x) /g^3 = -2(xxx +x +2x –2xxx) / g^3 =
= 2x(xx -3) / (xx+1)^3;