How do you use the first and second derivatives to sketch #y=e^x/x#?

Question

Paisley Johnson · Accepted Answer

graph{(e^x)/x [-22, 18, -9.88, 10.12]}

# y=e^x/x # to sketch the graph, we need to examine various behaviour of the function roots: # y =0 => e^x/x=0 # # :. e^x=0 # But as # e^x>0 AA x in RR =># no roots behaviour as #x->+-oo# As #x->-oo=>y->e^-oo/-oo->0^-# As #x->oo=>y->e^oo/oo->oo# asymptotes denominator#=0 => x=0# so a vertical asymptote when #x=0# turning (or critical points) #y=e^x/x#, so differentiating wrt #x# using the quotient rule #d/dx(u/v)=(v(du)/dx - u(dv)/dx)/v^2# gives: # dy/dx = (xd/dx(e^x)-e^xd/dx(x))/x^2 # # = (xe^x-e^x)/x^2 # # = ((x-1)e^x)/x^2 # At a critical point #dy/dx=0 => ((x-1)e^x)/x^2 = 0 # # :. (x-1)e^x=0 # # :. e^x=0# (no solutions); or #x-1=0=>x=1# When #x=1 => y=e^1/1=e# (or #~~2.7# ) so there is a critical point at #(1,e)# Nature of the critical points: We need to look at the second derivative; we can rearrange # dy/dx = ((x-1)e^x)/x^2 # as # dy/dx = e^x/x-e^e/x^2 # we already know the derivative of #e^x/x# (it is the above); so we must use the quotient rule again to find the derivative of #e^x/x^2# so we get # (d^2y)/dx^2 = ((x-1)e^x)/x^2-d/dx(e^x/x^2) # # :. (d^2y)/dx^2 = ((x-1)e^x)/x^2 - {(x^2d/dx(e^x) - e^xd/dx(x^2))/(x^2)^2} # # :. (d^2y)/dx^2 = ((x-1)e^x)/x^2 - {(x^2e^x - 2xe^x)/(x^4)} # That's quite a complex expression, so lets not even bother trying to simplify any more as we increase the chance of making a mistake; just substitute #x=1# to determine the nature of the turning point; #x=1 => (d^2y)/dx^2 = 0 - {(e^1 - 2e^1)/(1}}=e>0 =># minimum Sketching the graph: We now have enough to sketch the graph which actually looks like this; graph{(e^x)/x [-22, 18, -9.88, 10.12]}

Adam Adkins · Answer

undefined To sketch the function $ y = \frac{e^x}{x} $, we can use the first and second derivatives to analyze its behavior.

1. **First Derivative Analysis (Critical Points)**:
   - Find the first derivative of the function $ y' = \frac{d}{dx}\left(\frac{e^x}{x}\right) $.
   - Determine where $ y' = 0 $ or undefined. These are potential critical points.
   - Use these critical points to identify intervals where the function is increasing or decreasing.

2. **Second Derivative Analysis (Concavity)**:
   - Find the second derivative of the function $ y'' = \frac{d^2}{dx^2}\left(\frac{e^x}{x}\right) $.
   - Evaluate the sign of $ y'' $ in the intervals identified from the first derivative analysis.
   - Determine the concavity of the function in these intervals (concave up or down).

3. **Sketching the Graph**:
   - Plot the critical points and asymptotes (if any) on the coordinate plane.
   - Use the information from the first and second derivative analyses to sketch the behavior of the function between critical points.
   - Pay attention to the behavior of the function as $ x $ approaches positive and negative infinity.

4. **Optional Steps**:
   - You may want to calculate and plot additional points to get a more accurate sketch.
   - Consider the behavior of the function near any vertical asymptotes or points where the function is undefined.

By following these steps and carefully analyzing the first and second derivatives of $ y = \frac{e^x}{x} $, you can create a reasonably accurate sketch of the function.