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Suppose for a function f, that 1<= f'(x)<=3 show that

4<=f(6)-f(2)<=12

I don't even no where to start... I was thinkin that in some way you could use the mean value theorem!

f(c)= [f(6)-f(2)]/4

But I wouldn't have a clue where to go from there

2. Find the primitives of the functions

secxtanx

so I sed that u=secx and that du=secxtanx

So does this mean that the answer will just be equal to one?

The last one is just hard in the fact that I am unsure of wat u should be equal to...

integral x^6lnx(dx)