f(x)=x sqrt[(x^2)+16]?

Question

1

f(x)=x sqrt[(x^2)+16]
defined on [-7,4] interval

f(x) concave down on the region ? to ?
f(x) concave up on the region ? to ?
inflection point?
minimum?
maximum?

2.
f(x) = x^3/[(x^2)-4]

defined on the interval [-15,15]

vertical asymptotes (2 of them)??

concave up on the region ? to ?
and ? to ?

inflection point?

Answer 1

1. The funxction is concave down on those intervals where the second derivative is negative, and concave up where the second derivative is positive. An inflection point is a point where the curve changes concavity, either from + to - or from - to +.

To find min and max points, set the first derivative equal to zero;the values where the first derivative vanishes are critical values. Use the second derivative test to decide whether the critical values lead to max, or min, or neither.

2. The vertical asymptotes occur where the denominator is 0.

(You didn't expect me to do your homework for you, did you?)