To determine
To evaluate:
a) What happens to the maximum or minimum points and inflection points as c changes
b) Illustrate them by drawing graphs for different values of c.
Answer
a) Thus, x=12c is point of maxima and x=−12c is point of minima if c >0.
b) Thus, points x=±32c,0 are points of inflection if c >0.
Explanation
Given: Family of curves: f(x)=cxe−cx2 where c is real number.
Calculation:
Consider, the function
f(x)=cxe−cx2
(a) Maximum and minimum points:
Consider
f(x)=cxe−cx2
Differentiate it with respect to x, we get
f'(x)=ddx[cxe−cx2]=c[ddx(x)e−cx2+xddxe−cx2] (By product rule) =c[1×e−cx2+xe−cx2ddx(−cx2)] (By chain rule) =ce−cx2[1+x(−2cx)]=ce−cx2[1−2cx2]
Put f'(x)=0. This implies that
ce−cx2[1−2cx2]=0So, 1−2cx2=0 This gives thatx2=12c provided if c≠0⇒x=±12c provided if c>0.
Consider
f'(x)=ce−cx2[1−2cx2]
Differentiate it with respect to x, we get
f''(x)=ddx[ce−cx2[1−2cx2]]=c[ddx(e−cx2)(1−2cx2)+e−cx2ddx(1−2cx2)] (By product rule)=c[e−cx2ddx(−cx2)(1−2cx2)+e−cx2(0−4cx)] (By chain rule)=ce−cx2[−2cx(1−2cx2)−4cx]=ce−cx2[−2cx+4c2x3−4cx]=ce−cx2[4c2x3−6cx]=2c2e−cx2[2cx3−3x]=2c2e−cx2x[2cx2−3]
At x=12c
f''(12c)=2c2e−c×12c×12c[2c×12c−3] =2c2e−122c[1−3] =−4c2e−122c<0
Thus, x=12c is point of maxima provided c >0.
At x=−12c
f''(−12c)=2c2e−c×12c×−12c[2c×12c−3] =−2c2e−122c[1−3] =4c2e−122c>0
Thus, x=−12c is point of minima provided c >0.
Thus, x=12c is point of maxima and x=−12c is point of minima provided if c >0.
Inflection point:
Put f''(x)=0, we get
0=f''(x) =2c2e−cx2x[2cx2−3]This implies thatx[2cx2−3]=0 if c≠0⇒x=0 or x2=32c provided if c≠0which means x=0 or x=±32c provided if c>0.
Now, for all c>0
| Interval |
Sign of f'' on the interval |
Concave up or down |
| (−∞,−32c) |
−ve |
Concave down |
| (−32c,0) |
+ve |
Concave up |
| (0,32c) |
−ve |
Concave down |
| (32c,∞) |
+ve |
Concave up |
In this case x=±32c,0 are points of inflection.
Thus, points x=±32c,0 are points of inflection if c >0.
(b)
For different values of c some graphs are:
