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# 数学代写|微积分代写Calculus代考|Finding Absolute Extrema on a Closed Interval

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## 数学代写|微积分代写Calculus代考|Finding Absolute Extrema on a Closed Interval

Every function that’s continuous on a closed interval has an absolute maximum and an absolute minimum value in that interval a highest and lowest point – though, as you see in the following example, there can be a tie for highest or lowest value.

A closed interval like $[2,5]$ includes the endpoints 2 and 5. An open interval like $(2,5)$ excludes the endpoints.

Finding the absolute max and min is a snap. All you do is compute the critical numbers of the function in the given interval, determine the height of the function at each critical number, and then figure the height of the function at the two endpoints of the interval. The greatest of this set of heights is the absolute max; and the least is the absolute min. Example: Find the absolute max and min of $h(x)=\cos (2 x)-2 \sin x$ in the closed interval $\left[\frac{\pi}{2}, 2 \pi\right]$.

1. Find the critical numbers of $h$ in the open interval $\left(\frac{\pi}{2}, 2 \pi\right)$.
\begin{aligned} h(x) & =\cos (2 x)-2 \sin x \ h^{\prime}(x) & =-\sin (2 x) \cdot 2-2 \cos x \ 0 & =-2 \sin (2 x)-2 \cos x \ 0 & =\sin (2 x)+\cos x \ 0 & =2 \sin x \cos x+\cos x \ 0 & =\cos x(2 \sin x+1) \end{aligned}
(by the chain rule)
(now divide both sides by -2 )
(now use a trig identity)
(factor out $\cos x$ )
$$\begin{array}{rlrlrl} \cos x & =0 & \text { or } & & 2 \sin x+1 & =0 \ x=\frac{3 \pi}{2} & & \sin x & =-\frac{1}{2} \ & & x & =\frac{7 \pi}{6}, \frac{11 \pi}{6} \end{array}$$
Thus, the zeros of $h^{\prime}$ are $\frac{7 \pi}{6}, \frac{3 \pi}{2}$, and $\frac{11 \pi}{6}$, and because $h^{\prime}$ is defined for all input numbers, this is the complete list of critical numbers.
2. Compute the function values (the heights) at each critical number.
\begin{aligned} & h\left(\frac{7 \pi}{6}\right)=\cos \left(2 \cdot \frac{7 \pi}{6}\right)-2 \sin \left(\frac{7 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \ & h\left(\frac{3 \pi}{2}\right)=\cos \left(2 \cdot \frac{3 \pi}{2}\right)-2 \sin \left(\frac{3 \pi}{2}\right)=-1-2 \cdot(-1)=1 \ & h\left(\frac{11 \pi}{6}\right)=\cos \left(2 \cdot \frac{11 \pi}{6}\right)-2 \sin \left(\frac{11 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \end{aligned}
1. Determine the function values at the endpoints of the interval.
\begin{aligned} & h\left(\frac{\pi}{2}\right)=\cos \left(2 \cdot \frac{\pi}{2}\right)-2 \sin \left(\frac{\pi}{2}\right)=-1-2 \cdot 1=-3 \ & h(2 \pi)=\cos (2 \cdot 2 \pi)-2 \sin (2 \pi)=1-2 \cdot 0=1 \end{aligned}
So, from Steps 2 and 3, you’ve found five heights: 1.5, 1, 1.5, -3 , and 1 . The largest number in this list, 1.5 , is the absolute max; the smallest, -3 , is the absolute min.

The absolute max occurs at two points: $\left(\frac{7 \pi}{6}, 1.5\right)$ and $\left(\frac{11 \pi}{6}, 1.5\right)$. The absolute min occurs at one of the endpoints, $\left(\frac{\pi}{2},-3\right)$. Figure 6-7 shows the graph of $h$.

## 数学代写|微积分代写Calculus代考|Finding Absolute Extrema over a Function’s Entire Domain

A function’s absolute max and absolute min over its entire domain are the highest and lowest values of the function anywhere it’s defined. A function can have an absolute max or min or both or neither. For example, the parabola $y=x^2$ has an absolute min at the point $(0,0)$ – the bottom of its cup shape – but no absolute max because it goes up forever to the left and the right. You could say that its absolute max is infinity if it weren’t for the fact that infinity is not a number and thus it doesn’t qualify as a maximum (ditto, of course, for negative infinity as a minimum).

The basic idea is this: Either a function will max out somewhere or it will go up forever to infinity. And the same idea applies to a min and going down to negative infinity.

To locate a function’s absolute max and min over its domain, find the height of the function at each of its critical numbers – just like in the previous section, except that here you consider all the critical numbers, not just those in a given interval. The highest of these values is the absolute max unless the function rises to positive infinity somewhere, in which case you say that it has no absolute max. The lowest of these values is the absolute min, unless the function goes down to negative infinity, in which case it has no absolute min.

If a function goes up or down infinitely, it does so at its extreme right or left or at a vertical asymptote. So, your last step (after evaluating all the critical points) is to evaluate $\lim {x \rightarrow x} f(x)$ and $\lim {x \rightarrow-\infty} f(x)$ – the so-called end behavior of the function – and the limit of the function as $x$ approaches each vertical asymptote from the left and from the right. If any of these limits equals positive infinity, then the function has no absolute max; if none equals positive infinity, then the absolute max is the function value at the highest of the critical points. And if any of these limits is negative infinity, then the function has no absolute min; if none of them equals negative infinity, then the absolute min is the function value at the lowest of the critical points.

Figure 6-8 shows a couple functions where the above method won’t work. The function $f(x)$ has no absolute max despite the fact that it doesn’t go up to infinity. Its max isn’t 4 because it never gets to 4, and its max can’t be anything less than 4, like 3.999, because it gets higher than that, say 3.9999. The function $g(x)$ has no absolute min despite the fact that it doesn’t go down to negative infinity. Going left, $g(x)$ crawls along the horizontal asymptote at $y=0$. $g$ gets lower and lower, but it never gets as low as zero, so neither zero nor any other number can be the absolute min.

## 数学代写|微积分代写Calculus代考|Finding Absolute Extrema on a Closed Interval

\begin{aligned} h(x) & =\cos (2 x)-2 \sin x \ h^{\prime}(x) & =-\sin (2 x) \cdot 2-2 \cos x \ 0 & =-2 \sin (2 x)-2 \cos x \ 0 & =\sin (2 x)+\cos x \ 0 & =2 \sin x \cos x+\cos x \ 0 & =\cos x(2 \sin x+1) \end{aligned}
(根据链式法则)
(现在两边同时除以-2)
(现在用三角恒等式)
(提出$\cos x$)
$$\begin{array}{rlrlrl} \cos x & =0 & \text { or } & & 2 \sin x+1 & =0 \ x=\frac{3 \pi}{2} & & \sin x & =-\frac{1}{2} \ & & x & =\frac{7 \pi}{6}, \frac{11 \pi}{6} \end{array}$$

\begin{aligned} & h\left(\frac{7 \pi}{6}\right)=\cos \left(2 \cdot \frac{7 \pi}{6}\right)-2 \sin \left(\frac{7 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \ & h\left(\frac{3 \pi}{2}\right)=\cos \left(2 \cdot \frac{3 \pi}{2}\right)-2 \sin \left(\frac{3 \pi}{2}\right)=-1-2 \cdot(-1)=1 \ & h\left(\frac{11 \pi}{6}\right)=\cos \left(2 \cdot \frac{11 \pi}{6}\right)-2 \sin \left(\frac{11 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \end{aligned}

\begin{aligned} & h\left(\frac{\pi}{2}\right)=\cos \left(2 \cdot \frac{\pi}{2}\right)-2 \sin \left(\frac{\pi}{2}\right)=-1-2 \cdot 1=-3 \ & h(2 \pi)=\cos (2 \cdot 2 \pi)-2 \sin (2 \pi)=1-2 \cdot 0=1 \end{aligned}

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