# y= (x/2)^x  (Domain: x> 0) Analyze the functions using graph sketching techniques and then sketch the graph   Domain. restricting your attention only to values of x for which f(x) is defined....

1. y= (x/2)^x  (Domain: x> 0) Analyze the functions using graph sketching techniques and then sketch the graph

1. Domain. restricting your attention only to values of x for which f(x) is defined.
2. Intercepts. Find intersections of the graph with the axes. The Y-intercept is (0, f(0)). To obtain the X-intercept (c, 0) is harder and sometimes impossible. You need to find c such that f(c)= 0
3. Symmetry. If f(x)=f(-x) , for example cos x = cos (-x), then the graph is symmetric with respect to the Y-axis. If f(x)= -f(-x), for example sin x= - sin (-x), then the graph is symmetric with respect to the origin.
4. Intervals of increase or decrease and local maxima and minima Reminder: Solve f’(x)> 0 and f’(x)<0 for monotonicity

Sketch the curve Use all information gathered above to mark first points where the graph changes its behavior (intercepts, a minimum, a maximum, an inflection point) and then connect the points.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find x and y intercepts such that:

`x = 0 =gt f(0) = 0`

`y = 0 =gt f(x) = 0 =gtx = 0`

Hence, the graph of the function passes through the origin `(0,0).`

You need to check if the function is even or odd such that:

`f(-x) = (-x/2)^(-x) =gt f(-x) = 1/((-x/2)^x)`

Notice that the function is not odd, nor even.

You need to find the minimum and the maximum points, hence, you need to solve the equation f'(x) = 0 such that:

`f'(x) = ((x/2)^x)'`

`y = ((x/2)^x) =gt ln y = x ln (x/2)`

`(1/y)*y' = ln(x/2) + x(2/x)*(1/2)`

`(1/y)*y' = ln(x/2) + 1 =gt y' = y*(ln(x/2) + 1)`

`y' = ((x/2)^(x))*( ln(x/2) + 1)`

You need to solve f'(x) = 0 such that:

`((x/2)^(x))*( ln(x/2) + 1) = 0`

`((x/2)^(x)) = 0 =gt x = 0`

`ln(x/2) + 1 = 0 =gt ln(x/2)= -1 =gt x/2 = e^(-1) =gt x = 2/e`

Notice that the f'(x) is negative if `x in (0 ; 2/e)`  and it is positive if `x in (2/e ; oo).`

Hence the function reaches its minimum at `x = 2/e` .

Analyzing the monotony of the function yields that the function decreases over `(0 ; 2/e)`  and it increases over `(2/e ; oo).`

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