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You know that $f'(x) > 2f(x) > 2e^{2x}$ for every $x \in (0,\infty)$ which is incompatible with $f'(x) < e^{2x}$ for any $x \in (0,\infty)$.

Since the building has a square front its dimensions are $x \times x \times y$ where $x$ is both the height and width of the building and $y$ is its length. The areas of the front and back are $x^2$, …

Here is a hint: if each $a_n \le 1$, then $a_n^p \le a_n$ for all $n$.

Since $9^9 = 387420489$, you have $$\log_{10} 9^{9^9} = 9^9 \log_{10} 9 = 369693099.6\ldots$$ so that the resulting number has 369,693,099 $+$ 1 digits.

This isn't implicit differentiation. The original expression equals $3s^3 - s^2 + 3s$ if $s \not= 0$ and is undefined otherwise because the fraction has the form "$\frac 00$". Thus the derivative is …

Hint: $$f(n) = - \int_x^1 t^{1/n - 1} \, dt.$$ What happens to $t^{1/n - 1}$ as $n$ increases?

As to why, there must be a precise definition since otherwise it would not be possible to prove much about convergent sequences---they would fail to be a useful mathematical tool. As for the meaning, …

Let $g(x) = -x^3 - x$ and let $f(x) = -2x$. Both have a negative first derivative, but they intersect at three points. Edit: to follow up on Mario Carneiro's comment below: try to convince yourself t …

Two problems pop out right away: first, the integral of a function $f$ is not given by the formula $$\int\int_Df(x,y)dA = \sum_{i=1}^n\sum_{j=1}^m(M_{f_i}-m_{f_i})(\Delta x_i \Delta y_j). $$ Second $f …

Another way to approach the problem is using the intermediate value theorem for derivatives. You can show there exist $n$ distinct points $x_1,\ldots,x_n$ satisfying $\displaystyle \sum_{k=1}^n \frac{ …

Let $g(x) = f(x^2)$. The average rate of change of a function $f$ on an interval $[a,b]$ is $\dfrac{f(b) - f(a)}{b-a}$. The average rate of change of $f$ on the interval $[1,16]$ is $$\frac{f(16) …

According to the mean-value theorem you have $$f(1/2) - f(x) \le |f(1/2) - f(x)| \le M |1/2-x|$$ for all $x \in [0,1]$ so that $$f(1/2) \le M|1/2 - x| + f(x)$$ for all $x \in [0,1]$. Now integrate fr …

From Stewart's calculus: A point $P$ on a curve $y=f(x)$ is called an inflection point if $f$ is continuous there and the curve changes from concave upward to concave downward or from concave downward …

If you know that $$\lim_{x \to \infty} \left( 1 + \frac tx \right)^x = e^t$$ for any real $t$ you can use the particular choice $t = -1$ to find that $$\lim_{x \to \infty} \left(\frac{x-1}{x} \right)^ …

If $p > 0$ then the function $f(t) = t^p$ is increasing on $[0,\infty)$. You can prove this using the first derivative. Thus if $0 \le x \le 1$ then $0 \le x^p \le 1$. If $n > 1$ then $p - 1 > 0$ so …