# Find the linear approximation of f(x) = lnx at x = 1 and use it to estimate ln1.37 L(x) = ln 1.37 ~~my problem is with linear approximationswhat do I do if the slope has an x in it? & I'm...

Find the linear approximation of f(x) = lnx at x = 1 and use it to estimate ln1.37

L(x) =

ln 1.37 ~~

my problem is with linear approximations

what do I do if the slope has an x in it?

& I'm not trying to get 2 questions answered, but this is another problem with the same issue:

Let f(x) = x^3. The equation of the tangent line to f(x) at x=6 is y= ______

I got

m = 3x^2

y = 216

x = 6

so when solving for b I had 216 = 6(3x^2) + b

which means there's 2 variables ... what did I do wrong?

I'm so confused, I tried plugging x with the x, but I got the wrong answer ?

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Linear approximation is used to find approximate functions by linear functions which are closer to each other.

A approximation to a function f(x) is given by linear approximation of f'(x) when x=a using;

f'(x) = f(a)+f'(a)(x-a)

f(x) = lnx

f'(x) = 1/x

f'(1) = 1/1 =1

By linear approximation;

f'(1) = f(1)+f'(1)(x-1) = ln1+1(x-1) = (x-1)

When x is closer to 1 by linear approximation;

**lnx `~~` (x-1)**

**So ln1.37`~~` (1.37-1) = 0.37**

for your second question see the graph y=x^3

dy/dx is the tangent of y=x^3

You can see the tangent is changing when x is changing because of the curvature.

So it is obvious that you should have a function of x for dy/dx.