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(a) Find the intervals on which $ f $ is increasing or decreasing.(b) Find the local maximum and minimum values of $ f $.(c) Find the intervals of concavity and the inflection points.

$ f(x) = e^{2x} + e^{-x} $

(a) $f(x)=e^{2 x}+e^{-x} \Rightarrow f^{\prime}(x)=2 e^{2 x}-e^{-x}, f^{\prime}(x)>0 \Leftrightarrow 2 e^{2 x}>e^{-x} \Leftrightarrow e^{3 x}>\frac{1}{2} \Leftrightarrow 3 x>\ln \frac{1}{2} \Leftrightarrow$$x>\frac{1}{3}(\ln 1-\ln 2) \Leftrightarrow x>-\frac{1}{3} \ln 2[\approx-0.23]$ and $f^{\prime}(x)<0$ if $x<-\frac{1}{3} \ln 2 .$ So $f$ is increasing on $\left(-\frac{1}{3} \ln 2, \infty\right)$and $f$ is decreasing on $\left(-\infty,-\frac{1}{3} \ln 2\right)$(b) $f$ changes from decreasing to increasing at $x=-\frac{1}{3} \ln 2 .$ Thus,\[f\left(-\frac{1}{3} \ln 2\right)=f(\ln \sqrt[3]{1 / 2})=e^{2 \ln \sqrt[3]{1 / 2}}+e^{-\ln \sqrt[3]{1 / 2}}=e^{\ln \sqrt[3]{1 / 4}}+e^{\ln \sqrt[3]{2}}=\sqrt[3]{1 / 4}+\sqrt[3]{2}=2^{-2 / 3}+2^{1 / 3}[\approx 1.89]\]is a local minimum value.(c) $f^{\prime \prime}(x)=4 e^{2 x}+e^{-x}>0$ [the sum of two positive terms]. Thus, $f$ is concave upward on $(-\infty, \infty)$ and there is nopoint of inflection.

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Okay, so we're going to find the intervals, anguish, effort, increasing and decreasing the local mats. And then con Catherine inflection point of effort. So, in order to find their inter world on, whichever is increasing and decreasing we have to take were filled by the first of which means finding too derivative. So in this case will apply the chain rule to both eater to actually to the negative X overtake. The directive to act is just too. And then we take Stewart of E t. Rex and let's keep Two x gonna rewrite this on DH. Then I take the Druid of Negative X, which will be negative one on. Then you take he negative x And then we said this function is called zero, but we're going to actually simplify them further. This's I'm going to rewrite the same exact thing so you can see what I'm doing. I'm going to pull out and e to the negative X. I'm gonna pull out e to the negative X and then I'm going to put remember that if I'm going to pull out each of the negative X right this eating the negative X, I need to be careful with our experiment will hear us. You have to add an extra two eating to us. Because if I write, if I have to eat three x over eighty x one e x canceling two makes two eggs. So I have to write to eat in three acts minus one, um, and said this is zero, not six zero And then each of negative X had no real solution. So we ignore that. And now we'LL have to eat to the reactor minus one equals zero. So you add one on both sides and divide right too. You get either three active vehicles one half Now it is your national on those side and just give us three X equals natural one half and then we're divided by three. So we get natural log of one half divided by three. So now we're going. Teo, apply our sign Trying to this Wei are only going to consider the entire function of prime and we only evaluating it at one point, which is a very weird points of Ellen of one half over three. I'm going to put that over, bring a line here, and if you had to plug it in. Joe, you're talking a very big negative number into half time, which is right here. You get a r a negative number and very positive number. Get a positive number, isn't now we know the signs of the prime we know that is decreasing on negative Infinity two Ellen one half third and then increasing on Alan one half, One third, two part of infinity. And then, since this is a function that is decreasing in an increasing, we know that the local Max I mean local men occurs at act cool, Alan. One third to one half, one third and there's no local Max. And now we're one to be finding the con cavity and inflection point. We take the second derivative and we apply or change again. And if we apply a shingle again, we get for E two x plus e to the negative. And now, if you try to set, this equals zero. Um, you'll realise that one thing before we even set a decoded zero, you have to realize that this function, the exponential function it is they are all I'm sure you realize that this function of prime as double crime is actually positive. It is all positively appointment Any number of extra people in a very negative number. You still get it very put. You get a positive number plugging a negative number. You get a positive eating an egg. Correct. And this is still a positive number because an exponent functions so all values of X gives a positive. So that means that since it is always positive, always positive Ueno it is. Khan came up so negative was conking out from negative infinity to infinity. Akane came up and since there's no sign of current, no sign changes occurring there, not calm cavity. We know that there was no inflection point. So justice, I know when you're when you're working with accurate and function, you always want to look for slight clues, like if it is positive on all of the universe, because big metal rod did it evaluated could just skip the whole process of playing equals zero. So just decide now when you're looking at economic crime, should think more critically about the values of X and what sign they are and that is it

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