Tutor profile: Hungju W.

Explain the fundamental theorem of linear algebra. Fully comprehend each part of the theorem is crucial and essential to better understand linear algebra as a subject.

1. Firstly we establish the four fundamental subspaces given that $$ A $$ is an $$m \times n $$ matrix. These four subspaces are the range space of $$A $$ and the nullspace of $$A$$ and their transpose. $$R(A), R(A^T), N(A),N(A^T)$$ 2. $$dim(R(A)) = dim(R(A^T))$$ and $$dim(R(A)) + dim(N(A)) = n$$ 3. $$N(A) \bot R(A^T)$$

Solve for x the equation $$ ( \log(2+ \log_2 (x+1)) ) = 0$$

We can use the log properties to solve the equations. Firstly, we know that $$log_2 4 = 2$$ and we substitute 2 to get $$( \log(\log_2 4+ \log_2 (x+1)) ) = 0$$. Secondly, using the log property we combine the 2 items besides the plus sign and get $$( \log(\log_2 4(x+1)) ) = 0$$ Lastly, we know from the log property that $$\log_n 1 =0$$ so we know that $$(\log_2 4(x+1)) = 1$$, which means that $$ 4(x+1) =2, x=\frac{-1}{2}$$

Evaluate $$\frac{d}{dx} F(x) $$ given that $$ F(x) = \int_{2}^{3x} sin(t)dt$$.

We can utilize the second fundamental theorem of calculus together with chain rule.Since $$F(x)$$ has the form $$\int_{2}^{u(x)} f(t)dt $$, we can use chain rule $$\frac{d}{dx} F(x) = \frac{dF}{du} \cdot \frac{du}{dx} $$. We can use the second fundamental theorem to get the $$\frac{dF}{du}$$ part, which is $$\frac{dF}{du} = sin(u(x)) = sin(3x) $$. A simple chain rule can help us get the second part $$ \frac{du(x)}{dx} = 3 $$. So the answer will be $$\frac{d}{dx} F(x) = sin(3x) \cdot 3$$