Question
Consider the function below. (If you need to use - or , enter -INFINITY or INFINITY.) h(x) = (x + 2)^7 - 7x - 2 (a) Find the intervals of increase. (Enter the interval that contains smaller numbers first.) ( , ) ( , ) Find the interval of decrease. ( , ) (b) Find the local minimum value. Find the local maximum value. (c) Find the inflection point. ( , ) Find the interval the function is concave up. ( , ) Find the interval the function is concave down. ( , ) (d) Use this information to sketch the graph of the function. (Do this on paper. Your instructor may ask you to turn in this graph.)
Explanation / Answer
Let f(x) = y. Hence 1 + 5x - 7x^2 = y This is a quadratic equation. The curve of f(x) is a U shaped curve since the coeficient of the x^2 term is neagtive (-7). (a) The curve of f(x) is continuous. As x increases from 0 then f(x) tends to -oo. Likewise as x decreases from 0, f(x) tends to -oo. Since -oo, which, is not a number the curve of f(x) has no vertcal asymptotes. The curve will show a maximum (see later) but again the curve is continuous and does not have any horizontal asymptotes. (c) Let's tackle (c) first: Differentiate (I assume we can use calculus?) f(x) = 1 + 5x - 7x^2 Hence dy/dx = 5 - 7x At any turning points ( local maxima, local minima or points of inflexion) dy/dx = 0. Hence set: 5 - 7x = 0 => 7x = 5 => x = 5/7 Hence f(x) has a turning point at x = 5/7. We need to show this is either a minumum, a maximum or a turning point. (note, we have to show this and not rely on our knowledge of quadratic curves which suggests it will be a maximum) To do this we use the second derivative test. dy/dx = 5 - 7x => d^2y/dx^2 = -7 which is < 0 and hence the point is a local maximum. To find the y-coordinate substitute for x in f(x). This gives y = 1 Hence we conclude that the curve of f(x) has a local maximum at x = 5/7, y = 1. That is at point (5/7,1).A lso, the curve does not have any minima or points of inflexion. (b) Now we can consider (b). Since the curve of f(x) has no vertical or horizontal asymptotes and a local maximum at x = 5/7 we need to consider the nature of the slope on the intervals: (-oo, 5/7) and (5/7, oo) To do this just evaluate dt/dx at a convenient values of x in each interval. Thus: When x = 0, dy/dx = 5 which is > 0 and hence the slope is increasing. When x = 1, dy/dx = -2 which is < 0 and hence the slope is increasing. This is what we would expect with the curve having a maximum at x = 5/7 but we have now showen it to be the case. hence we conclude that: f(x) is decreasing on the interval (-oo, 5.7) and increasing on the interval (5/7, oo) (d) Finally, from the above we can deduce that the curve is concave up on the interval (-oo, oo). Of course this excludes it being concave down on any interval. Note, the description here is what the curve looks like in a normal x-y coordinate system.