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The figure shows a beam of length $L$ embedded in concretewalls. If a constant load $W$ is distributed evenly along itslength, the beam takes the shape of the deflection curve$$y=-\frac{W}{24 E I} x^{4}+\frac{W L}{12 E I} x^{3}-\frac{W L^{2}}{24 E I} x^{2}$$where $E$ and $I$ are positive constants. $(E$ is Young's modulusof elasticity and $I$ is the moment of inertia of a cross-sectionof the beam.) Sketch the graph of the deflection curve.

$x=\frac{6 L \pm \sqrt{12 L^{2}}}{12}=\frac{1}{2} L \pm \frac{\sqrt{3}}{6} L$

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 4

Curve Sketching

Derivatives

Differentiation

Applications of the Derivative

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So we're wanting to use the guidelines of this section in order to find the graph of a seemingly complicated formula. And that's going to be F of X equals negative W over 24 E I X. to the 4th plus W L Over 12 E.I X cubed minus W L squared Over 24 e X squared. Um And we could factor this out. We can assume that the domain is anywhere from zero to out. So the first thing we want to do is find the X intercept. And why intercept? So for X intercept, Remember we want to let we want to find F of zero. We want to find value and Y equals zero. So we end up getting that that zero. Um and then we check for anywhere else. So the other X intercept is going to be zero and X equals L. And then, because of the domain, we don't need to test for f of negative acts. So that's part C. For part D. We see that the domain is a closed interval. So we don't really need to check the limits at infinity because it's closed. So then E and F for E. We want to check the derivative graph, we have primary equals zero. And when we do that, we saw for X and get X equals zero. Then using that we uh see that X when X equals zero, We also get the x equals L over two. And al. So we're also going to look at double prime in order to make things a little simpler for us. And we see that it is concave up at the critical number L over two. So that is going to be our local minimum. And based on that, we see that we have from zero to well over two is decreasing. And then from L over 2, 2, L that's going to be increasing. And then finding where F. Prime of zero or F double prime equals zero. We get that X equals three. L. A. Plus or minus route three. L over six. So with that we have our different intervals to concave up in concave down ones. So we start with concave down that's going to be zero to this right here, that's concave down. And then from this value two this volume. So we have to keep in mind that we have the plus and minuses. So this one right here is gonna be minus and then this one right here is gonna be from minus to plus. Um That's gonna be concave up, this one's concave down. And then lastly we have from this guy right here positive to L. Is going to be concave down. Um So then if we plot the graph, what we end up getting is something that looks like this um for our final graph

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