Graph of the function y=(x−2)e−x.
The graph is drawn below.
Consider, the function
(1) Domain of the function y:
Domain of ex is ℝ. Therefore,
Domain of y=(x−2)e−x is also ℝ.
For y-intercept, put x = 0 in the function y:
Therefore, (0,−2) is y-intercept. For x-intercept put y(x)=0. This gives
This implies that
Thus, x =2 and so x-intercept is the point (2, 0).
This implies that y is neither even nor odd function.
Since the function y is finite for all values, therefore, it has no vertical asymptote.
limx→∞y(x)=limx→∞(x−2)e−x=limx→∞x−2ex (∞∞-form)=limx→∞1ex (By l'Hospital's rule)=0
Thus, y =0 is horizontal asymptotes.
(5) Increasing or decreasing:
Differentiate it with respect to x, we get
y'=ddx[(x−2)e−x]=ddx(x−2)e−x+(x−2)ddxe−x (By product rule)=1×e−x+(x−2)×e−xddx(−x) (By chain rule) =e−x+(x−2)×e−x(−1) =e−x(1−x+2)=e−x(3−x)
Put y'(x)=0. This implies that
e−x(3−x)=0So, 3−x=0 as exponential function is always positive.This gives thatx=3.
This implies that y is increasing in the interval (−∞,3) and decreasing in (3,∞).
(6) Local maxima or minima:
Therefore, x=3 is point of maxima.
(7) Inflection point:
Put y''(x)=0, we get
0=y''(x) =e−x[x−4]This implies that(x−4)=0⇒x=4
Therefore, in the intervals (−∞,4) function is concave down and on the interval (4,∞) it is concave up.
Thus, by the above information’s graph of y is:
The above graph is on small scale.
In the large scale the graph is visible like this: