0
votes
3answers
50 views

Where did the linear approximation/linearization formula come from?

Where did the linear approximation/linearization formula come from? I understand that it takes root in the point-slope form and slope intercept form of a linear equation, but I don't understand where ...
0
votes
1answer
25 views

Use economisation to find linear approximation to x^2-x-1?

I've been given the solution to this question... It uses chebychev, and you get: $1/2(2x^2-1)-2x-1/2$ So the Chebyshev economisation polynomial is $-2T1 -1/2 T0$ I can see the logic in how this ...
1
vote
1answer
25 views

How do I do Linearization at a point that lies on a curve?

I keep applying the formula to the info given but I keep getting lost/weird answers. Can someone please help me? I know $L(x)=f(a)+f'(a)(x-a)$ question Y(x) satisfies $x^2y^2 + xy = 6$. Point (x,y) ...
1
vote
2answers
68 views

Linear approximation formulas for $x$ not near $0$

I am following MIT's calculus videos and I have noticed that when dealing with linear approximations, the professor calculates a set of approximation "formulas" for $x$ near $0$ like $1+x$ for $e^x$ , ...
1
vote
2answers
915 views

Linear approximation to ln(x) at x = 1, then estimate ln(1.08)

I know that the derivative of $\ln(x)$, or log of whatever base (x) = $(1/x)$ *the original function. If x is a more complicated expression, then the derivative would be $(x'/x)*f(x)$. If I knew the ...
0
votes
3answers
1k views

Linear approximation to $y = \sqrt{1-x}$ at $x=0$, then approximate $\sqrt{0.9}$ and $\sqrt{0.99}$

How do I find this? I know that the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$. Here, I would plug in $(1-x)$ instead of $x$. When $x = 0$, the slope would evaluate to $\dfrac{1}{2}$. I got ...
0
votes
1answer
67 views

Which number is larger if f'(x) is a differentiable increasing function for all x?

Suppose is a differentiable increasing function for all x. Which number is the larger and why? or ? I believe f(x) must be concave up everywhere since the derivative is increasing, but I am not ...
0
votes
1answer
328 views

Relative error; absolute error divided by real value or approx value?

I have the following function; $$f(x) = -\frac{x}{2x + 4}\cdot v_r \Rightarrow \hat{f}(x) = -\frac{x}{4}\cdot v_r$$ Because $x$ is very small we can approximate $f(x)$ to $\hat{f}(x)$. Now the ...