The graph of a function is concave up on an interval.?

Question

The graph of a function is concave up on an interval. Describe the behaviour of the first derivative and the second derivative of this function on the interval.

Answer 1

We presume that the interval lies in first quadrant, so the first derivative describes the rate of change of quantity, Ty (along y-axis) w.r.t. quantity Tx(along x-axis, in other words describes the slope of the curve. slope is also given by tangent at any point on the curve. since the curve is concave up . we so if we start with low value of x, the slope or the tangent makes an obtuse angle with the x- axis hence is negative and it goes on increasing as we increas x till it becomes zero(point of minimum at the curve) and then becomes positive. In other words Ty first decreases with increasing Tx , becomes zero, and then start increasing with increasing Ty.

Second derivative on the other hand describes how the first derivative is changing. since the first derivative has an increasing value throughout the interval, so it is evident that the second derivative will also increase and will be positive in this case.

Answer 2

The second derivative is positive on this interval. The first derivative is increasing.

Answer 3

f(x)=(x-a million)/(x+3) f'(x) = [(x+3)-(x-a million)]/(x+3)^2 f'(x) = (x+3-x+a million)/(x+3)^2 f'(x) = 4/(x+3)^2 f'(x) = 4(x+3)^(-2) f''(x) = 4(-2)(x+3)^(-3) = -8/(x+3)^3 f is concave up if f''(x) > 0 -8/(x+3)^3 > 0 8/(x+3)^3 < 0 the answer is x < -3 f is concave up on (-00, -3).