Master Differential Calculus Problems with Our Experts

Find the inverse function if:
$2e^{3x}=4e^{5x}$

Find the inverse function if:
$9^x=2e^{x^2}$

If a ball is throwndirectly upward with a velocity of 40ft/s, its height (in feet)after t seconds is given by $y=40t-<!--\; -\; -->16t^2$. What is the maximum heightattained by the ball?

Show that a curve has constant speed if and only if its acceleration is everywhere orthogonal to its velocity.

For the scalar function $T=e^{-<!--\; -\; -->r/5}\cos <!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}$ , determine its directional derivative along the radial direction $\stackrel{^<!--\; ^\; -->}{r}$ and then evaluate it at $P(2,\mathrm{\pi <!--\; \pi \; -->}/4.3)$.

Determine the nature and stability of thecritical point (0,0) for the following system:
$\frac{dx}{dt}=-<!--\; -\; -->\sin <!--\; \; -->(x-<!--\; -\; -->y)$
$\frac{dy}{dt}=1-<!--\; -\; -->5y-<!--\; -\; -->e^x$

Determine the total number of critical points of the function $f(x)=(x+e^x{)}^k$, where $k>0$ is an integer

Find the derivative $\frac{d}{dx}[\arctan <!--\; \; -->(x^2+1)]$
$$\begin{gathered} (arcsec(x^2+1){)}^2 \\ \frac{1}{2+x^2} \\ \frac{1}{1+(x^2+{)}^2} \\ \frac{2x}{1+(x^2+1{)}^2} \end{gathered}$$

Suppose $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ has two continuous derivatives, has only one critical point $x_0$, and that $f^{″}(x_0)<0$. Then $f$ achieves its global maximum at $x_0$, that is $f(x)\le <!--\; \le \; -->f(x_0)$ for all $x\in <!--\; \in \; -->\mathbb{R}$.

In order to find the minimum/ maximum point of a function, I would take it's derivative and find critical points, points where the derivative is zero or undefined, and then put them on a number line and do strawberry field. If it's negative to the left of the point and positive to the right of the point, the point is a minimum, etc. When the denominator of your fraction is equal to zero, isn't that where you have a vertical tangent line and not a candidate for a min/max point?

what is the derivative of $(\ln <!--\; \; -->(x){)}^x$

Whats the difference between the critical point of a function and the turning point?Aren't they both just max/min points?

I am preparing for my exam and need help with the following task:
Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a continuous function with the estimation $x^2\le <!--\; \le \; -->f(x)$$\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$. Show that f takes on its absolute minima.
If a function is continuous, then $\underset{x\to <!--\; \to \; -->a}{lim}f(x)$=$f(a)$.
Our function has an absolute minima in $x_0$$\in <!--\; \in \; -->\mathbb{R}$, if $f(x)\ge <!--\; \ge \; -->f(x_0)$ for $x\in <!--\; \in \; -->\mathbb{R}$.
At first I thought the task is pretty easy. We learned how to prove that if $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous, then f has an absolute maxima and an absolute Minima in [a,b]. The Problem here is, that the domain of our function here is unbounded. Thats why I don't have any idea what I could and should use for the proof. We should probably use the estimation $x^2\le <!--\; \le \; -->f(x)$ $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$. This gives us the information, that f is bounded from below with $f(x)\ge <!--\; \ge \; -->0$ $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$. But how does this help me? And what else do we have?

Define $f:D(0,1)-<!--\; -\; -->>C$ to be holomorphic such that $f(0)=0$. I want to extend $\frac{f(z)}{z}$ to be continuous on $\stackrel{¯<!--\; ¯\; -->}{D}(0,r)$ for arbitrary $0<r<1$.
My initial guess was to define:
$g(z)=\frac{f(z)}{z}$ for non-zero $z$ but extend $g(z)$ to be 0 for $z=0$.
But if I take $\underset{z\to <!--\; \to \; -->0}{lim}\frac{f(z)}{z}=f^{\prime }(0)$ using L'Hopital's rule. So I think this should be the expression for $g(0)$? I wonder if:
i) L'hopital's rule is applicable to complex functions, and,
ii)Am I taking the limit incorrectly? I also do not see a use for $f(0)=0$ in my deduction so I'm rather perplexed.
Any help would be appreciated!

Prove: If $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_n=L$ and $a_n>a$ for all n then $L\ge <!--\; \ge \; -->a$

Explain in your own words what the Intermediate Value Theorem says and why it seems plausible.
.A.
The Intermediate Value Theorem states that for a continuous function​ f(x) over the closed interval​ [a,b], f(x) takes on every value between​ f(a) and​ f(b). This seems plausible because if the function did not take on every value between​ f(a) and​ f(b), there would be a value for​ f(x) that did not​ exist, and thus the function would not be continuous.
B
The Intermediate Value Theorem states that for every value of x between a and b in​ f(x), if​ f(a) exists, then​ f(b) must exist if b is in the domain. This seems plausible because if the function did not take on every value between​ f(a) and​ f(b), there would be a value for​ f(x) that did not​ exist, and thus the function would not be continuous.
C.
The Intermediate Value Theorem states that for a continuous function​ f(x) over the closed interval​ [a,b], f(x) takes on every value between​ f(a) and​ f(b). This seems plausible because if b is in the domain of​ x, then​ f(b) must exist.
D.
The Intermediate Value Theorem states that for every value of x between a and b in​ f(x), if​ f(a) exists, then​ f(b) must exist if b is in the domain. This seems plausible because if b is in the domain of​ x, then​ f(b) must exist.

Find the critical point of
$f(x)=\frac{5x}{x-<!--\; -\; -->3}$

What is the basic idea for finding critical point via Morse theory and critical groups?

solve the following limit without L'Hospital rule or series:
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{\arctan <!--\; \; -->2x}{\sin <!--\; \; -->3x}$