Showing Nonexistence of Maxima, Minima, etc.

[turophile]

Here's the problem:

Show that, if ad ? bc, f(x) = (ax + b)/(cx + d) has no local maxima, no local minima, and no inflection points. Sketch the graph for a = c = 1, b = – 1, d = 0.

I'm able to show that there are no local maxima or minima:

f = (ax + b)/(cx + d)
f' = [(cx + d) ? a – (ax + b) ? c]/(cx + d)[sup:1h0irh2d]2[/sup:1h0irh2d] = (acx + ad – acx – bc)/(cx + d)[sup:1h0irh2d]2[/sup:1h0irh2d] = (ad – bc)/(cx + d)[sup:1h0irh2d]2[/sup:1h0irh2d]
f'' = [(cx + d)[sup:1h0irh2d]2[/sup:1h0irh2d] ? 0 – (ad – bc) ? 2 ? (cx + d) ? c]/(cx + d)[sup:1h0irh2d]4[/sup:1h0irh2d] = 2c(bc – ad)(cx + d)/(cx + d)[sup:1h0irh2d]4[/sup:1h0irh2d] = 2c(bc – ad)/(cx + d)[sup:1h0irh2d]3[/sup:1h0irh2d]

Since ad ? bc, f' is never zero, so there are no local minima or maxima.

Now I'm not sure how to show that there is no inflection point. I can tell from the graph the problem asks for that the function is concave up for x < 0 and concave down for x > 0, but also that it is not continuous at x = 0. So although evidently there is a change of sign in f'' from x < 0 to x > 0, there is no inflection point because x is not defined at 0. My question: How do I show this observation from the graph using math? It seems that f'' could be equal to 0 (and therefore there may be a point of inflection) if c = 0 and d ? 0. I can't think of what the next step should be.

 

[Deleted member 4993]

Here's the problem:

Show that, if ad ? bc, f(x) = (ax + b)/(cx + d) has no local maxima, no local minima, and no inflection points. Sketch the graph for a = c = 1, b = – 1, d = 0.

I'm able to show that there are no local maxima or minima:

f = (ax + b)/(cx + d)
f' = [(cx + d) ? a – (ax + b) ? c]/(cx + d)[sup:199qcpn4]2[/sup:199qcpn4] = (acx + ad – acx – bc)/(cx + d)[sup:199qcpn4]2[/sup:199qcpn4] = (ad – bc)/(cx + d)[sup:199qcpn4]2[/sup:199qcpn4]
f'' = [(cx + d)[sup:199qcpn4]2[/sup:199qcpn4] ? 0 – (ad – bc) ? 2 ? (cx + d) ? c]/(cx + d)[sup:199qcpn4]4[/sup:199qcpn4] = 2c(bc – ad)(cx + d)/(cx + d)[sup:199qcpn4]4[/sup:199qcpn4] = 2c(bc – ad)/(cx + d)[sup:199qcpn4]3[/sup:199qcpn4]

Since ad ? bc, f' is never zero, so there are no local minima or maxima.

Now I'm not sure how to show that there is no inflection point. I can tell from the graph the problem asks for that the function is concave up for x < 0 and concave down for x > 0, but also that it is not continuous at x = 0. So although evidently there is a change of sign in f'' from x < 0 to x > 0, there is no inflection point because x is not defined at 0. My question: How do I show this observation from the graph using math? It seems that f'' could be equal to 0 (and therefore there may be a point of inflection) if c = 0 and d ? 0. I can't think of what the next step should be.

If c = 0

f = a/d x + b/d -- a straight line and hence no inflection point.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \frac{ax+b}{cx+d}, \ and \ we \ are \ given \ that \ a=c=1, \ b=-1, \ and \ d \ = \ 0, \ hence \ ad \ \ne \ bc.\)

\(\displaystyle Now \ f(x) \ = \ 1-\frac{1}{x} \ \implies \ f'(x) \ = \ \frac{1}{x^2} \ = \ 0, \ impossible, \ 1 \ \ne \ 0, \ hence \ no\)

\(\displaystyle critical \ points \ \ and \ when \ x \ = \ 0, \ f(x) \ is \ undefined, \ so \ no \ critical \ points \ there.\)

\(\displaystyle f"(x) \ = \ \frac{-2}{x^3} \ = \ 0, \ impossible, \ -2 \ \ne \ 0, \ hence \ no \ points \ of \ reflection.\)

\(\displaystyle Note: \ when \ f(x) \ = \ \frac{ax+b}{cx+d}, \ if \ we \ set \ f'(x) \ and \ f"(x) \ = \ to \ 0, \ we \ get \ ad \ = \ bc,\)

\(\displaystyle but, \ we \ are \ given \ that \ ad \ \ne \ bc, \ hence \ no \ critical \ points \ or \ points \ of \ inflection.\)

\(\displaystyle See \ graph \ below\)

[attachment=0:388bf4t5]hhh.jpg[/attachment:388bf4t5]