For all real values of x, the minimum value of is A mathematical expression representing the fraction (1 minus x plus x squared) divided by (1 plus x plus x squared).

Q28 of 168 Page 231

For all real values of x, the minimum value of is

Let

Then, f’(x) = 0

⇒ x² = 1

⇒ x = �1

Now,

And,

Also, f’’(-1) = -4 < 0

Then, by second derivative test, f is minimum at x = 1 and the minimum value is given by

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is .

The point on the curve x² = 2y which is nearest to the point (0, 5) is

The maximum value of is

Using differentials, find the approximate value of each of the following: