Even Differentiable Functions and Linearization

[n310]

Is there anything special about even differentiable function of x? Give reasons behind your answer.
and
Find the linearization of
g(x)= 3+ ∫sec(t-1)dt at x=-1
It is a definite intergral going from 1 to x^2.. a=1 b=x^2

I understand how to do regular linearization problems but with this having a definite integral I am having a hard time.
With the first question the most I can get out of it is that every even differentiable function ends with an odd function. In order words its symmetric to the y-axis. However, I need more reasons.
I would appreciate the help as soon as possible.
Thank You

 

[HallsofIvy]

I have no idea what you mean by "every even differentiable ends with an odd function". In what sense does one function "end" with another?

It is true that the graph of an even function is symmetric to the y-axis. What does that tell you about its derivative (the slope of the tangent line to the graph) at x= 0?

The "linearization" of a function, f(x), at x= a, is f(a)+ f'(a)(x- a). Here they are referring to the linearization at x= -1 which is f(-1)+ f'(-1)x.

 

[n310]

I know the formula for Linearization. I just did not know how to solve for it considering it had a definite integral.
I am confused with the first question that's why I asked for more input.
Thank You for the reply :)

 

[lanedance]

ok, well g is a function of x, so u will be linearising w.r.t. x, and x appears in the upper limit of the definite integral

can you try and differentiate the integral?

though a little sloppy, it may help to assume you know the indefinite integral, call it F(t), then apply the limits and differentiate

 

[n310]

Natural Logarithms

a) Prove that f(x)=x-ln x is increasing from x>1.
b) Using part (a), show that lnx < x if x>1

I don't know how to go about this to start it off, I would appreciate the help, and a few steps to get me started asap. Thank You

 

[n310]

thank you, I got a bit of help from my classmates, took a while but I got there..

 

[lanedance]

no worries, if you have new questions, please make a new post