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Handout Monday 9/15
A brief note about section 1.7. As we have discussed, functions might not have a derivative everywhere. Section 1.7 looks at various ways that limits might not exist, and since derivatives are limits, these might not exist either.
But actually, speaking of this, consider what happens when you zoom in at a point that is differentiable vs one that is not. Not being differentiable usually means that is a “cusp” at that point (although points with vertical tangent lines are also not differentiable).
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Look at graphs on desmos. Maybe \(|x-2|+1\text{.}\) No matter how far we zoom in, we still have a cusp.
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But now look at \((x-2)^2 + 1\text{.}\) Zoom in at the point \((2,1)\) and you see that if you zoom in enough, the curve looks like a horizontal line.
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In fact, also graph the tangent line, \(y = 1\text{.}\) Can you tell these apart when you zoom in? No!
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A differentiable function and its tangent line share both a common point and a common derivative at that point.
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Since it is often easier to evaluate the equation of a line at a point than a function at a point, we can approximate a function at a point by evaluating the tangent line there.
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So two important, closely related ideas are:
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Given a function \(f\) and a point \((a,f(a))\text{,}\) we can consider a line \(y - f(a) = f'(a)(x - a)\) that also passes through the point \((a,f(a))\) and has slope equal to \(f'(a)\text{.}\)
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Given a function \(f\) and a point \((a,f(a))\text{,}\) we can consider a linearization function \(L(x) = f'(a)(x-a)+f(a)\) that satisfies \(L(a) = f(a)\) and \(L'(a) = f'(a)\text{.}\)
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The only difference between these two is the context we are thinking about. The first is a set of points that satisfies an equation. The second is a linear function (whose graph is exactly that line).
Here is an example. What is \(\sqrt{4.1}\text{?}\) Okay, we know that \(\sqrt{4}\) is 2. In other words, think of \(f(x) = \sqrt{x}\) and we are asking for \(f(4.1)\text{.}\) What is \(f'(x)\text{?}\) We could evaluate a limit and find \(f'(x) = \frac{1}{2\sqrt{x}}\text{.}\) Now find \(L(x)\) for \(a = 4\text{.}\) We get \(L(x) = \frac{1}{4}(x-4) + 2\)
So what is \(L(4.1)\text{?}\) That’s pretty easy to compute: \(L(4.1) = 0.25(4.1-4) + 2 = 0.025 + 2 = 2.025\text{.}\) Since the local linearization is a good approximation for the function for values of \(x\) close to \(4\text{,}\) we have that \(f(4.1) = \sqrt{4.1} \approx 2.025\text{.}\) (In fact, \(f(4.1) = 2.02485\)).
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In section 1.5, we used this idea too. Suppose \(f(t)\) is the temperature \(t\) hours after midnight. \(f(6) = 58\) and \(f'(6) = -2\text{.}\) Can we approximate \(f(6.25)\text{?}\)
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Those statements tell us that at 3am, the temperature is 58 degrees, and is dropping at a rate of 2 degrees per hour. I don’t know what is going to happen in the next hour (maybe the temperature will start increasing), but our best bet is that in the next 15 minutes (0.25 hours), it will decreasing 1/2 a degree.
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We can write down the local linearization function: \(L(x) = -2(x-6)+58\text{.}\) Evaluate this at \(6.25\text{:}\) \(L(6.25) = -2(.25)+58 = 57.5\text{.}\)
