That's all they give you. I tried putting it into the linear approximation equation of:  $$ f(a)+f'(a)(x-a) $$  but I get almost the same value as $64.07^{2/3}$, which is around $16.0116$. Just not sure how else to go about doing this problem. Thanks.  Here's the problem. I'm doing this online and it comes back as incorrect. I've tried nearly every possible solution to this (16.0117, 16.0116, 16.012, 16, 16.1, etc.) Is there a specific solution to this? Thanks.
122281
That's all they give you. I tried putting it into the linear approximation equation of:  $$ f(a)+f'(a)(x-a) $$  but I get almost the same value as $64.07^{2/3}$, which is around $16.0116$. Just not sure how else to go about doing this problem. Thanks.
24533
Please help, I'm stuck in this problem:     Find the values of $a$ and $b$:   $$\frac{a(26.273)^b}{a(27.165)^b}=1.0065809$$
32456
I've got an equation that contains   $$x^p - 1$$   $x$ is any positive number (such as 2) and $p$ is a small positive number close to 0 (such as 0.001).  For some reason (that I may have known in High School), this can be approximated by   $$\ln(x) \times p$$  Can anyone explain why this approximation works?  Does it have something to do with approximating a natural log with a series?
2201
I want a to be rounded to 13.95.  &gt;&gt;&gt; a 13.949999999999999 &gt;&gt;&gt; round(a, 2) 13.949999999999999   The  function does not work the way I expected.
77630
Consider $f\in C^2$ so that $$f''(x)+f(x)=-x\,g(x)f'(x), \ \forall x\ge0 $$ where $g(x)\ge 0$. Then ($\forall x\ge 0$)         (A) $f(x)^2+f'(x)^2$ is non-increasing,   (B) $f(x)^2&lt;3f(0)^2+(2f'(0))^2$,   (C) $|f(x)|\le\alpha$, where $\alpha$ is a fixed real constant.   (D) $\lim_{x\to\infty} f(x)\sin\left(\dfrac{1}{x}\right)$ exists.        Answer Given: (A),(B),(C),(D)  What I have tried : I differentiated option (A) and then with the help of the given equation in the question I was able to show that  it is less than zero . Hence I was able to infer that option (A) is correct .   Now coming to option (B).  I assumed the function in option (A) to be $h(x)$.   $$ h(x) = f(x)^2 + f'(x)^2 $$  Since $h'(x)&lt;0$ ,therefore $h(x)$ is a decreasing function   hence for all $x\ge 0$  $h(x)&lt; h(0)$  which implies $(f(x))^2 + (f'(x))^2&lt;(f(0))^2 + (f'(0))^2$  From the above expression we can get the second option .   Is my method correct ?. If not , please show the right method .  I am unable to get the the 3rd and 4th option .
1631
I got this problem from my teacher as a optional challenge. I am open about this being a given problem, however it is not homework. The problem is stated as follows. Assume we have an infinite tetration as follows $$x^{x^{x^{.^{.^.}}}} \, = \, a$$ With a given $a$ find $x$. The next part of the problem was to discuss the convergence radius of a. If a is too big or too small the tetration does not converge. Below is my humble stab at the problem.  My friend said you would have to treat the tetration as a infinite series, and therefore could not perform algebraic manipulations on it before it is know whether it converges or diverges. However my attempt is to first do some algebraic steps, then discuss the convergence radius. I) Initial discussion At the start it is obvious that the tetration converges when $a=1$ (just set $x=1$) Now after some computer hardwork it seems that the tetration fails to converge when a is roughly larger than 3. II) Algebraic manipulation $$ x^{x^{x^{.^{.^.}}}} \, = \, a$$ This is the same as $$ x^a \, = \, a$$ $$ \log_a(x^a) \, = \, \log_a(a)$$ $$ \log_a(x) \, = \, \frac{1}{a}$$ $$ x \, = \, a^{\frac{1}{a}}$$ Now, if we let $a=2$ then $x = \sqrt{2}$. After some more computational work, this seems to be correct, which makes me believe this formula is correct. III) Discsussion about convergence By looking at the derivative of $ \displaystyle \large a^{\frac{1}{a}} $ we see that the maxima occurs when $a=e$, which also seems to correspond to the inital computational work. Now I think, that the minima of $\displaystyle \large a^{1/a}$ is zero by looking at its graph, studying its derivative and the endpoints. So that my &quot;guess&quot; or work shows that it converges when $$ a \in [0 \, , \,  1/e] $$ VI) My questions Can my algebraic manipulations be justified? They seem rather sketchy taking the a`th logarithm and so on . (Although they seem to &quot;magically&quot; give out the right answer) By looking at Wikipedia it seems that the tetration converge when $$ a \in \left[ 1/e \, , \, e \right] $$ This is almost what I have, why is my lower bound wrong? How can  I find the correct lower bound?
99168
Let $$l=\lim_{x \rightarrow a}\frac{x^x-a^x}{x-a} $$&amp;$$m= \lim_{x \rightarrow a}\frac{a^x-x^a}{x-a} $$ where $a&gt;0$. If $l=m$ then find the value of $'a'$. I get $$l=a^alna  $$ $$m= a^alna-a^a $$  Certainly not correct as $a&gt;0$ ?
31513
You can't use L'Hospital's rule :S  $$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$$  I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
47599
The minimum value of $$f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$$ is  (a)-1  (b)9  (c)6  (d)1  Apart from trying to obtain $1$, which in this case is simple and $f(2)=1$ is there a standard method to approach such problems.  Please keep in mind that this is an objective question in one of the competitive exam and you get around 5 mins to solve it.  Also this is asked in elementary section, so only knowledge of basic calculus and polynomials is assumed.
9060
I'm having some difficulties continuing this problem. I get that $b^2\mid a$ and $c\mid a$ but I am not sure where to go from there.
53224
I am a 12th grade student, and I am afraid that in realistic terms this question might not even make sense because of the infinities that have to be dealt with. However, in my attempt to calculate it's derivative, I did the following:  $$y=x^{x^{x^{.^{.^.}}}}$$  $$\ln(y)=\ln(x)x^{x^{x^{.^{.^.}}}}$$  $$\ln(y)=y\ln(x)$$  after taking derivative with respect to x on both sides, I obtained the following:  $$\frac{\mathrm dy}{\mathrm dx}=\frac{y^2}{x(1-\ln(y))}$$  I am fairly certain that the calculations up to this point are valid. However, to further continue, I analyzed the nature of y in different domains and obtained the values, by observation(that is, by observing how $x$, $x^x$, $x^{x^x}$, $x^{x^{x^x}}$, and so on would behave to draw a conclusion):   for $x&lt;1,$ $y$ becomes 0 for $x=1,$ $y$ becomes 1 for $x&gt;1,$ $y$ becomes infinite   These 3 points is where the first problem lies. Are these true? If so, how do we reach to the conclusion?  Secondly, considering this to be true, I get the derivative at:   $x&lt;1$ to be 0. $x=1$ to be 1. for $x&gt;1$, I took $x=2.$ then the derivative $\frac{\mathrm dy}{\mathrm dx} = y^2/[2(1-ln(y))]$ (replacing $x$ by $2$). Now, I applied L hospital's rule to get the value of the expression to be negative infinity.   This is the second problem. I have used L Hospital's rule, but limits were not concerned. Is this method valid? If not, how would we calculate it?
78855
Can someone please explain me what is a ?  Also, how does putting $$X_a(\psi) ~=~ x_a\star\psi$$ realise $$[X_a,X_b]=i\theta_{ab}{\bf 1}?$$   Ref:
107160
I'm hoping someone could check my work. I'm trying to simplify the equation for $x$. $$(\frac{x}{a})^{3.2} + (\frac{y}{b})^{3.2} = 1$$ $$\frac{a}{b} = \frac{174.1}{86}$$ Thus I began by substituting $a$ with: $$a = \frac{174.1}{86}b$$ $$((\frac{86}{174.1})(\frac{x}{b}))^{3.2} + (\frac{y}{b})^{3.2} = 1$$ $$((\frac{86}{174.1})(\frac{x}{b}))^{3.2} = 1 - (\frac{y}{b})^{3.2}$$ $$(\frac{x}{b})^{3.2} = \frac{(1 - (\frac{y}{b})^{3.2})(174.1)^{3.2}}{(86)^{3.2}}$$ $$x^{3.2} = \frac{(1 - (\frac{y}{b})^{3.2})(174.1)^{3.2}(b)^{3.2}}{(86)^{3.2}}$$ $$x^{3.2} = \frac{(1 - (\frac{y}{b})^{3.2})((174.1)(b))^{3.2}}{(86)^{3.2}}$$ $$x^{\frac{16}{5}} = \frac{(1 - (\frac{y}{b})^{3.2})((174.1)(b))^{3.2}}{(86)^{3.2}}$$ $$x = [\frac{(1 - (\frac{y}{b})^{3.2})((174.1)(b))^{3.2}}{(86)^{3.2}}]^{\frac{5}{16}}$$ Is my work correct, or am I completely off the board?
98334
Give $f (x) = x^3 - 4x^2 - 3x + 18$  For which values of $x$, will $f (x) . f'(x) &lt; 0$?  And also, how can I understand these type of questions? I always get stuck on them. Other examples are:  $$\frac{f(x)}{f'(x)} &gt; 0$$ $$xf (x) &gt; 0$$
78581
I used the equation   $$k = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}.$$  Just by looking at its would think the max is at $x = 1, y = 0$.  What is the correct answer?
12544
Let  $$f(x,y)=\frac{x^3-y^3}{x^2+y^2}$$  My solution manual says that this function is not diffb. in $(0,0)$ because it is not linear. Well my problem is that I don't see why this function is linear, and I also don't see why that would imply that $f$ is not differentiable.
34545
I have tried to let$f(a)=\int_0^1\frac{\ln(1-ax)}{x} dx$, which means     $f'(x)=-a\int_0^1 \frac{1}{x(1-ax)} dx=a\int_0^1 \left[\frac{a}{ax-1}-\frac{1}{x}\right] dx$.   However, I am stuck at here.