p. 473

70.

f[x_]  :=  (x - 3)^2 E^x

Plot[f[x], {x, -1, 4}]

f'[x]

Simplify[%]

Solve[3 - 4 x + (x^2)  ==  0]

f''[1]

f''[3]

Plot[f'[x], {x, -1, 4}]

We see f (x)  >  0 for x  <  1 and x  >  3, so f (x)is increasig on those intevals; and f' (x)  <  0 for 1  <  x  <  3 so f (x) is decreasing on this interval. The above also shows that f (x) has a relative maximum at x = 0 and a minimum at x  =  3.

f''[x]

Simplify[%]

Plot[f''[x], {x, -1, 4}]

Solve[-1 - 2 x + (x^2) == 0]

We see that ({{f'' (x)  < , 0}}) for 1 - (Sqrt[2])  <  x  <  1  +  (Sqrt[2]), so f (x) is concave down there;
outside this interval f'' (x)  >  0, so f (x)is concave up. The points 1 - (Sqrt[2])and  1  +  (Sqrt[2])are thus inflection points.