Tangent lines and natural logs homework help.

[chris40256]

Homework Statement

1. Let f be the function defined by f(x) = -2 + ln(x^2).
a) For what real numbers x is f defined
b) Find the zeros of f
c) Write an equation for the line tangent to the graph of f at x=1

Homework Equations

The Attempt at a Solution

a) all positive reals except 0 ( i believe)
b) Teacher let's us use calculator so : 2.718 , -2.718 (no hurt in checking that)
c)at x = 1 , y = -2
Finding slope/derivative:
0+ (1/x^2)(2x)
2x/(x^2)
plug 1 in for the slope: 2(1) / (1)^2
m= 2
Equation: y+2 = 2(x-1) (i believe its correct but I'm not sure)

 

[rock.freak667]

a) Should be correct as it is given in that form of ln(x^2)
b) -2 +ln(x^2)=0
ln(x^2)=2
x^2=e^2

I think you forgot to take the sq. root after

c) Correct

 

[HallsofIvy]

Homework Statement

1. Let f be the function defined by f(x) = -2 + ln(x^2).
a) For what real numbers x is f defined
b) Find the zeros of f
c) Write an equation for the line tangent to the graph of f at x=1

Homework Equations

The Attempt at a Solution

a) all positive reals except 0 ( i believe)
b) Teacher let's us use calculator so : 2.718 , -2.718 (no hurt in checking that)

Being allowed to use a calculator doesn't mean you are required to! No, 2.718 and -2.718 are NOT zeros of f. e and -e are. Do you understand the difference?
Oh, and whether your teacher requires it or not you should always show HOW you solve a problem, not just give the answer.

c)at x = 1 , y = -2
Finding slope/derivative:
0+ (1/x^2)(2x)
2x/(x^2)
plug 1 in for the slope: 2(1) / (1)^2

Are you serious? You actually put x= 1 into 2x/x^2? 2x/x^2= 2/x obviously. In fact, the whole problem is simpler if you write f(x)= -2+ ln(x^2)= -2+ 2ln(x). Then f' (x)= 2/x.

m= 2
Equation: y+2 = 2(x-1) (i believe its correct but I'm not sure)