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Numerade Educator



Problem 59 Hard Difficulty

(a) Use a graph of $ f $ to estimate the maximum and minimum values. Then find the exact values.
(b) Estimate the value of $ x $ at which $ f $ increases most rapidly. Then find the exact value.

$ f(x) = \frac{x + 1}{\sqrt{x^2 + 1}} $


&f(x)=\frac{x+1}{\sqrt{x^{2}+1}} \Rightarrow f^{\prime}(x)=\frac{1-x}{\left(x^{2}+1\right)^{3 / 2}}\\
&f^{\prime}(x)=0 \Leftrightarrow x=1 . f(1)=\frac{2}{\sqrt{2}}=\sqrt{2} \text { is the exact value }
\end{aligned}$$ (b) From the graph in part (a), $f$ increases most rapidly somewhere between $x=-\frac{1}{2}$ and $x=-\frac{1}{4}$. To find the exact value,
we need to find the maximum value of $f^{\prime},$ which we can do by finding the critical numbers of $f^{\prime}$
$f^{\prime \prime}(x)=\frac{2 x^{2}-3 x-1}{\left(x^{2}+1\right)^{5 / 2}}=0 \Leftrightarrow x=\frac{3 \pm \sqrt{17}}{4}, \quad x=\frac{3+\sqrt{17}}{4}$ corresponds to the minimum value of $f^{\prime}$
The maximum value of $f^{\prime}$ occurs at $x=\frac{3-\sqrt{17}}{4} \approx-0.28$


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Video Transcript

Okay, so we're being used, asked to use the graph of after estimate a maximum and minimum value. And then we're being asked to find the exact values for these, um, stir. So I'm looking at this function. We cannot democracy that there's some sort of local max occurring right here. And we don't see any men know Ben. So there is no men, so we're going to just ignore this case, eh? So we're gonna look at the local max. So we first got to do is you can take the derivative so we first find a crime and giant crowd of X is equal to experts. One over the square root of X squared. I'm No, I'm so sorry, everybody. The derivative is one minus x over X square plus one, two to five, half power. And what we look for is Wenders to every physical zero. Because when it equals, the other is a horizontal tangent line of zero. And that's where the maximum occurred. So we set this equal to zero and one minus X equals zero. That just X is equal to one. So we know that the local max is occurring at one and it makes sense. This is it looks like a somewhere around one. And if you plug in f of won into our function, just our regular function, we get rude too. And that is the value of the maximum value. So for the then, we're being asked to estimate the value effects at which F increases most rapidly. So this is a little bit more different question. So, uh, you asked me from, Basically wears a slope essentially is the steepest. That's that's basically what we're looking for. And so this is the steepest slope occurs generally around at the inflection point because that's when the khan cavity is changing and the contract changes has usually has a steep slope current witted on the function. So yes, you have to take the second derivative. So we'LL take a double crime and after the crime has comes out to be two x squared minus three x minus one all over X, where plus one two to five power and he said this is equal to zero sent to explore minister in France. One is a quadratic function that would not necessarily have a simple factor solution. You have to apply the quadratic laments that the negative B plus or minus squared B squared minus four a. C. I'm gonna go and write that down. So you do you apply this formula and there could be touch the line in the square root of B squared minutes for a C all over two. A. And Dr Done that already and you can't X equals our three plus or minus rude seventeen. Oh, for And so what we're going to do now is we're just going to plug in both values into a prime to figure out which locals steepest. So usually this will be done on a calculator because this is that much more complicated inputs. So just come down, you get a value of negative zero point zero nine and then if you plug in three of minus, who had seventeen over before you get one point one four and one point one four is obviously bigger than negative zero point zero nine said. I mean, this is the steepest point in our function, and that is it