1. Find (f-1)(a) for the function f and real number a.
f(x)=x[sup:3o0ay4v4]2[/sup:3o0ay4v4]-4/x, a=6
So far I have found that (f[sup:3o0ay4v4]-1[/sup:3o0ay4v4])(2)=6
The thing I’m not sure of is in this step:
f’(x)=3x[sup:3o0ay4v4]2[/sup:3o0ay4v4]+4/x[sup:3o0ay4v4]2[/sup:3o0ay4v4]
Why would the x change to x[sup:3o0ay4v4]2[/sup:3o0ay4v4] and why is it added?
2. Show that the slopes of the graphs of f and f-1 are reciprocals at the indicated point.
f(x)=x3 (1/2, 1/8)
f[sup:3o0ay4v4]-1[/sup:3o0ay4v4](x)=3?x (1/8, 1/2)
f’(x)=3x[sup:3o0ay4v4]2[/sup:3o0ay4v4]
f’(x)=3(1/2)[sup:3o0ay4v4]2[/sup:3o0ay4v4]=3/4
f’-1(x)= 1/3 [sup:3o0ay4v4]3[/sup:3o0ay4v4]?x[sup:3o0ay4v4]2[/sup:3o0ay4v4]for this part, why does the 3 go in front of the square root yet the 3 remains as a subscript on the square root and the x becomes x[sup:3o0ay4v4]2[/sup:3o0ay4v4]?
f(x)=x[sup:3o0ay4v4]2[/sup:3o0ay4v4]-4/x, a=6
So far I have found that (f[sup:3o0ay4v4]-1[/sup:3o0ay4v4])(2)=6
The thing I’m not sure of is in this step:
f’(x)=3x[sup:3o0ay4v4]2[/sup:3o0ay4v4]+4/x[sup:3o0ay4v4]2[/sup:3o0ay4v4]
Why would the x change to x[sup:3o0ay4v4]2[/sup:3o0ay4v4] and why is it added?
2. Show that the slopes of the graphs of f and f-1 are reciprocals at the indicated point.
f(x)=x3 (1/2, 1/8)
f[sup:3o0ay4v4]-1[/sup:3o0ay4v4](x)=3?x (1/8, 1/2)
f’(x)=3x[sup:3o0ay4v4]2[/sup:3o0ay4v4]
f’(x)=3(1/2)[sup:3o0ay4v4]2[/sup:3o0ay4v4]=3/4
f’-1(x)= 1/3 [sup:3o0ay4v4]3[/sup:3o0ay4v4]?x[sup:3o0ay4v4]2[/sup:3o0ay4v4]for this part, why does the 3 go in front of the square root yet the 3 remains as a subscript on the square root and the x becomes x[sup:3o0ay4v4]2[/sup:3o0ay4v4]?
