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
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= ______
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 ?
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.