# What are the values and types of the critical points, if any, of f(x)=(2x^2+5x+5)/(x+1)?

Feb 2, 2016

critical points are 0, -2, -1
Minima at (0,5), Maxima at (-2,-3), Discontinuous at x=-1

#### Explanation:

critical points are those at which f'(x) is =0, or for which f'(x) is not defined.
In this case f'(x)= $\frac{\left(4 x + 5\right) \left(x + 1\right) - \left(2 {x}^{2} + 5 x + 5\right)}{x + 1} ^ 2$

= $\frac{2 {x}^{2} + 4 x}{x + 1} ^ 2$

=$\frac{2 x \left(x + 2\right)}{x + 1} ^ 2$

Critical points are therefore x=0, x=-2 and x= -1

For determining the types find f"(x)= $\frac{\left(4 x + 4\right) {\left(x + 1\right)}^{2} - 2 \left(x + 1\right) \left(2 {x}^{2} + 4 x\right)}{x + 1} ^ 4$

For x=0, f"(x) would be +ive, hence it is a minima at the point (0, 5)

For x= -2, f"(x) would be -ive, hence it is a maxima at the point (-2,-3)

At x=-1, f(x) does not exist. x=-1 is a vertical asymptote