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Find the local maximum and minimum values of $f$ using both the First and Second Derivative Tests. Which method do you prefer?$f(x) = \sqrt{x} - \sqrt[4]{x}$

$f(x)=\sqrt{x}-\sqrt[4]{x} \Rightarrow f^{\prime}(x)=\frac{1}{2} x^{-1 / 2}-\frac{1}{4} x^{-3 / 4}=\frac{1}{4} x^{-3 / 4}\left(2 x^{1 / 4}-1\right)=\frac{2 \sqrt[4]{x}-1}{4 \sqrt[4]{x^{3}}}$First Derivative Test: $2 \sqrt[4]{x}-1>0 \Rightarrow x>\frac{1}{16},$ so $f^{\prime}(x)>0 \Rightarrow x>\frac{1}{16}$ and $f^{\prime}(x)<0 \Rightarrow 0<x<\frac{1}{16}$since $f^{\prime}$ changes from negative to positive at $x=\frac{1}{16}, f\left(\frac{1}{16}\right)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$ is a local minimum value.Second Derivative Test: $f^{\prime \prime}(x)=-\frac{1}{4} x^{-3 / 2}+\frac{3}{16} x^{-7 / 4}=-\frac{1}{4 \sqrt{x^{3}}}+\frac{3}{16 \sqrt[4]{x^{7}}}$$f^{\prime}(x)=0 \Leftrightarrow x=\frac{1}{16} . \quad f^{\prime \prime}\left(\frac{1}{16}\right)=-16+24=8>0 \Rightarrow f\left(\frac{1}{16}\right)=-\frac{1}{4}$ is a local minimum value.Preference: The First Derivative Test may be slightly easier to apply in this case.

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Catherine R.

Missouri State University

Heather Z.

Oregon State University

Michael J.

Idaho State University

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so to use the first derivative at the first derivative test to figure out the local minimum maximum values of our function F of X. First thing we need to do is find are derivatives. So F prime of X is equal to using the power rule one half X to the negative one half minus 1/4 X. to the -3/4. And I found this just using the power go like I said. And now what we want to do is just set this equal to zero in solve for X. So if we got zero equal one half X to the negative 1/2 -1 4 X to the -3/4. We can first ad 1/4 X to the negative 3/4 over to the other side. So you get 1/4 It's negative 3/4 is equal to have 1/2 X to the negative one half. And now we can multiply both sides by X. to the 3/4. And this will get rid of this X to the negative 3/4. So you can cancel these and we can add the exponents on this side of our exes. So negative one half plus 3/4 is 1/4 So we get 1/4 is equal to one half, X to the 1/4. And now we can multiply by two. So you get one half is equal to X to the 1/4. And now we're going to raise both sides to the 4th power and we get X is equal to one half to the fourth, which is 1 16, so are derivative is equal to zero at X is equal to 1/16. So what we want to look at our values of X that are less than 1/16 in values of X that are greater than 1/16 and look at the sign of our derivative to figure out if this is a max or minimum value. So if we go back to our derivative again and if we plug in a value of, let's just say one over 100 X to the negative one half of one over 100 would be one over 100 to the one half, which is 1/10. And then to the negative one would be 10. So you'd have 10 times one half which is five minus 1/4 one over 100 to the -3/4. So I'm just going to plug this into a calculator real quick and one over 100 to the three divided by four is equal to a brown .3. And then if we multiply that um by four we're sorry, we actually need to Take this to the -1. So we need to do one divided by Our answers, one divided by answer and then times 0.25 it's around 7.9. So we figured out one half, times one over 100 to the negative one half was five and we're minus ng seven. So that's going to give us a negative value for our derivative. So when X is less than 1/16, We have f prime of X being less than zero. And now we're just gonna look at values of exeter greater than 1/16. So let's just choose X is equal to one. So one to the negative, one half is just one. So we get one half minus 1/4 times one to the negative 3/4 which again, one to the negative 3/4 is just one. So this is one half minus 1/4 which is positive. So when X is greater than 1/16 are dead Are derivative, f prime of X is greater than zero. And since we're going from negative to positive or from a decreasing slope to increasing slope, that means that we have a local minimum at this X value. So at X is equal to 1 16 is a local minimum value. And now what we can do is just plug in 116 into our original equation. And if we do that and you do the math and we calculate this, this is equal to -1/4. So our local minimum occurs at 1/16 -1/4. And now if we want to use the second derivative test, so second derivative test to find um the local extreme instead, what we would do is find the second derivative of our function. So We know that f prime of X is equal to 1/2 X. to the negative 1/2 -1 4 X. A negative 3/4. So we can use the power rule again to find the second derivative. So F double prime of X. It's equal to one half times negative one half is negative 1/4 X. To the negative three halves minus 1/4 times negative 3/4 is negative 3/16. And then we also have this negative sign here. So this is positive 3/16 X. To the negative 7/4. So now that we've found this derivative And we know that our first derivative is equal to zero at X is equal to 1/16. What we can do is plug in 1/16 into our second derivative. So f double prime of 1/16. It's equal to negative 1 4th 1/16 to the -3/2, Plus 3/16 times 1/16 To the -7/4. And if you find this by plugging it into your calculator, this is going to be equal to eight which is greater than zero. And whenever our second derivative is greater than zero at a critical point, that means that that is a local minimum. So using the second derivative test, we found that this was a local minimum At the .1 16 comma negative 1/4. And in this case I preferred using the first derivative test since the second derivative was a little harder to find since we had some some weird numbers here. So I prefer using the first derivative test, but both work and both are great ways to figure out if you have a maximum or minimum value.

Oregon State University

Catherine R.

Missouri State University

Heather Z.

Oregon State University

Michael J.

Idaho State University

Lectures

Join Bootcamp