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# 数学代写|微积分代写Calculus代考|The First Derivative Test

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## 数学代写|微积分代写Calculus代考|The First Derivative Test

The First Derivative Test is based on the Nobel-prize-caliber ideas that as you go over the top of a hill, first you go up and then you go down, and that when you drive into and out of a valley, you go down and then up. This calculus stuff is pretty amazing, isn’t it?
Take a number line and put down the critical numbers you found above: 0, -2, and 2. See Figure 6-3.

This number line is now divided into four regions: to the left of -2 , from -2 to 0 , from 0 to 2 , and to the right of 2 . Pick a value from each region, plug it into the derivative, and note whether your result is positive or negative. Let’s use the numbers $-3,-1,1$, and 3 to test the regions.
\begin{aligned} f^{\prime}(x) & =15 x^4-60 x^2 \ f^{\prime}(-3) & =15(-3)^4-60(-3)^2=15 \cdot 81-60 \cdot 9=675 \ f^{\prime}(-1) & =15(-1)^4-60(-1)^2=15-60=-45 \ f^{\prime}(1) & =15(1)^4-60(1)^2=15-60=-45 \ f^{\prime}(3) & =15(3)^4-60(3)^2=15 \cdot 81-60 \cdot 9=675 \end{aligned}
These four results are, respectively, positive, negative, negative, or positive. Now, take your number line, mark each region with the appropriate positive or negative sign, and indicate whether the function is increasing (derivative is positive) or decreasing (derivative is negative). The result is a sign graph. See Figure 6-4.

Figure 6-4 simply tells you what you already know from the graph of $f-$ that the function goes up until -2 , down from -2 to 0 , further down from 0 to 2 , and up again from 2 on.

The function switches from increasing to decreasing at -2 ; you go up to -2 and then down. So at -2 you have a hill or a local maximum. Conversely, because the function switches from decreasing to increasing at 2, you have a valley there or a local minimum. And because the signs of the first derivative don’t switch at zero, there’s neither a min nor a max at that $x$-value (you get an inflection point when this happens).

The last step is to obtain the function values (heights) of these two local extrema by plugging the $x$-values into the original function:
$$\begin{gathered} f(-2)=3(-2)^5-20(-2)^3=64 \ f(2)=3(2)^5-20(2)^3=-64 \end{gathered}$$
The local $\max$ is at $(-2,64)$ and the local $\min$ is at $(-2,64)$.

## 数学代写|微积分代写Calculus代考|The Second Derivative Test

The Second Derivative Test is based on two more prize-winning ideas: That at the crest of a hill, a road has a hump shape, so it’s curving down or concave down; and that at the bottom of a valley, a road is cup-shaped, so it’s curving up or concave up.

The concavity of a function at a point is given by its second derivative: A positive second derivative means the function is concave up, a negative second derivative means the function is concave down, and a second derivative of zero is inconclusive (the function could be concave up, concave down, or there could be an inflection point there). So, for our function $f$, all you have to do is find its second derivative and then plug in the critical numbers you found $(-2,0$, and 2$)$ and note whether your results are positive, negative, or zero. To wit –

\begin{aligned} & f(x)=3 x^5-20 x^3 \ & f^{\prime}(x)=15 x^4-60 x^2 \quad \text { (power rule) } \ & f^{\prime \prime}(x)=60 x^3-120 x \quad \text { (power rule) } \ & f^{\prime \prime}(-2)=60(-2)^3-120(-2)=-240 \ & f^{\prime \prime}(0)=60(0)^3-120(0)=0 \ & f^{\prime \prime}(2)=60(2)^3-120(2)=240 \end{aligned}
At -2 , the second derivative is negative $(-240)$. This tells you that $f$ is concave down where $x$ equals -2 , and therefore that there’s a local max at $x=-2$. The second derivative is positive (240) where $x$ is 2 , so $f$ is concave up and thus there’s a local min at $x=2$. Because the second derivative equals zero at $x=0$, the Second Derivative Test fails – it tells you nothing about the concavity at $x=0$ or whether there’s a local min or max there. When this happens, you have to use the First Derivative Test.

## 数学代写|微积分代写Calculus代考|The First Derivative Test

\begin{aligned} f^{\prime}(x) & =15 x^4-60 x^2 \ f^{\prime}(-3) & =15(-3)^4-60(-3)^2=15 \cdot 81-60 \cdot 9=675 \ f^{\prime}(-1) & =15(-1)^4-60(-1)^2=15-60=-45 \ f^{\prime}(1) & =15(1)^4-60(1)^2=15-60=-45 \ f^{\prime}(3) & =15(3)^4-60(3)^2=15 \cdot 81-60 \cdot 9=675 \end{aligned}

$$\begin{gathered} f(-2)=3(-2)^5-20(-2)^3=64 \ f(2)=3(2)^5-20(2)^3=-64 \end{gathered}$$

## 数学代写|微积分代写Calculus代考|The Second Derivative Test

\begin{aligned} & f(x)=3 x^5-20 x^3 \ & f^{\prime}(x)=15 x^4-60 x^2 \quad \text { (power rule) } \ & f^{\prime \prime}(x)=60 x^3-120 x \quad \text { (power rule) } \ & f^{\prime \prime}(-2)=60(-2)^3-120(-2)=-240 \ & f^{\prime \prime}(0)=60(0)^3-120(0)=0 \ & f^{\prime \prime}(2)=60(2)^3-120(2)=240 \end{aligned}

## MATLAB代写

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