Answer to Question #92257 in Calculus for Fayoo

Question

Find the maximum value of the function f(x) = x + x2

Answer 1

Answer: there are no local maximum points.

$$supf(x)=+\infty$$

Explanation. The function f(x) is unbounded above, since

$$\lim_{x\to \infty }[(x^{2}+x+\frac{1}{4})-\frac{1}{4}]=\lim_{x\to\infty}[(x+\frac{1}{2})^{2}-\frac{1}{4}]=+\infty$$

The only stationary point

$$x=-\frac{1}{2} , (f'(-\frac{1}{2})=0)$$

is the local minimum point, as can be seen from the inequality:

$$f(x)=x+x^{2}=(x+\frac{1}{2})^{2}-\frac{1}{4}\geqslant -\frac{1}{4}=f(-\frac{1}{2})$$

There are no maximum points