Name: Rain Xia       Student Number: 59118159

Use the function, derivative and second derivative to sketch the graph of f(x) =x^x

f(x)=x^x     domain: x≥0.      intercepts: when x=0, y=1, therefore the intercept is (0,1).                                asymptotes: lim x->∞f(x)=∞      lim x->0 f(x)=1

f'(x)=(x^x)(1+lnx)     intervals of increase and decrease: when (x^x)(1+lnx)<0.368, the function decreases, when (x^x)(1+lnx)>0.368, the function increases.     extrema: f'(x)=(x^x)(1+lnx)=0 when x=0.368, therefore the extrema for f(x) is x=0.368 and it is the global minimum of the funtion.

f”(x)=x^x*[(lnx+1)²+1/x]     intervals of concavity: x^x*[(lnx+1)²+1/x]≥0, therefore f(x) is concave up.     inflection points: (0.368,0.692)

Based on these points and information, we can draw the diagram of function f(x)=x^x.

Name: Rain Xia       Student Number: 59118159

There is a fountain in UBC campus which has a 5 meters radius and 1 meter height. If water flows into the fountain at a rate of 5 meters^3 per minute, how fast is the depth of the water increasing?

Given: dv/dt=5

Want: dh/dt

V=πr^2h

dv/dt=π*5^2*dh/dt

5=π*25*dh/dt

dh/dt=1/(5π)

Therefore the speed of the water increasing is 1/(5π) meter per minute.