Question

59 - 62. sketching curves sketch a graph of a function f that is continuous on (-∞, ∞) and has the following properties.

1. ( f(x) < 0 ) and ( f(x) > 0 ) on ( (-infty, 0) ); ( f(x) < 0 ) and ( f(x) < 0 ) on ( (0, infty) )

Explanation:

Step1: Analyze the first - derivative

Since \(f^{\prime}(x)<0\) on \((-\infty,\infty)\), the function \(f(x)\) is decreasing on the entire real line.

Step2: Analyze the second - derivative

On \((-\infty,0)\), \(f^{\prime\prime}(x)>0\), so the function is concave up on \((-\infty,0)\). On \((0,\infty)\), \(f^{\prime\prime}(x)<0\), so the function is concave down on \((0,\infty)\).

Answer:

Start with a point. Since the function is decreasing everywhere (\(f^{\prime}(x)<0\)), as \(x\) increases, \(y = f(x)\) decreases. For \(x<0\), the curve is concave up (like \(y=-x^{2}\) for \(x < 0\) but decreasing). For \(x>0\), the curve is concave down (like \(y =-x^{2}\) reflected in some way but still decreasing). A possible function is \(y=-e^{-|x|}\). The graph has a horizontal asymptote (as \(x\to\pm\infty,y\to0\)). It is decreasing for all \(x\). For \(x < 0\), \(y^{\prime}=e^{-x}>0\) (wait no, let's use the rules of derivatives. Let \(f(x)=-\frac{1}{x^{2}+1}\) (adjusted). Or a better - constructed function: \(f(x)=-\int_{0}^{x}\frac{1}{1 + t^{2}}dt-\frac{1}{2}x\). But a simple sketch: Draw a curve that is decreasing. For \(x<0\), it bends upwards (concave up) and for \(x > 0\), it bends downwards (concave down). For example, start at a high \(y\) - value when \(x\) is very negative. As \(x\) approaches \(0\) from the left, the function is decreasing but the slope of the tangent (which is \(f^{\prime}(x)\)) is getting less negative (since \(f^{\prime\prime}(x)>0\) for \(x < 0\)). At \(x = 0\), there is a point of inflection. As \(x\) moves to the right of \(0\), the function continues to decrease but now the slope of the tangent is getting more negative (since \(f^{\prime\prime}(x)<0\) for \(x>0\)).