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MermaidWonders

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In summary: Let's look at an example that more clearly demonstrates the statement regarding concavity isn't always true. Consider:H(x)=\ln(x+1)And so:H'(x)=h(x)=\frac{1}{x+1}We have:h(t)>0 for $0\le t\le1$But:H''(x)=-(x+1)^{-2}On the given interval, this function is concave down, but it does not meet the given criteria, so we know the statement is...true. :)

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MermaidWonders

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MarkFL

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Suppose $h(t)=c$, where $0<c$...what is $H(x)$?

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MermaidWonders

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If $$c$$ is a constant, then $$H(x)$$ = $$cx$$?

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tkhunny

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What is the definitive test for "Concave Up"?

Can you apply this test to your function?

Can you apply this test to your function?

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MermaidWonders

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Like where a function's second derivative is > 0?

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Greg

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For inline latex, use \$...\$.

For example, \$x^2=4\$ gives $x^2=4$, inline.

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MermaidWonders said:Like where a function's second derivative is > 0?

Yep.

From wiki:

f is called

If f is twice-differentiable, then f is

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MermaidWonders said:Wait, so how do I get the sign of $H''(x)$? From the question, I know that since the fundamental theorem tells me that $H'(x)$ = $h(x)$, and that $h(t) > 0$, $H'(x)$ will also be $< 0$. But how do I figure out the concavity from there?

Indeed. So we know that $H'(x)>0$ (positive not negative) for $0\le x \le 1$, but we still don't know anything about $H''(x)$.

We can only tell that $H$ is strictly increasing and that it is strictly positive on the interval. We cannot say if it's convex or concave.

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MarkFL

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I like Serena said:Indeed. So we know that $H'(x)>0$ (positive not negative) for $0\le x \le 1$, but we still don't know anything about $H''(x)$.

We can only tell that $H$ is strictly increasing and that it is strictly positive on the interval. We cannot say if it's convex or concave.

Indeed, and is why I resorted to looking for a function satisfying the given criteria, that is not concave up on the given interval. :)

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MarkFL said:Indeed, and is why I resorted to looking for a function satisfying the given criteria, that is not concave up on the given interval. :)

Unfortunately, with $h(t)=c$, the function $H$ is actually concave up ($H''(x)\ge 0$), just not

But yes, it does illustrate that we cannot just make the function (strictly) concave up.

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MarkFL

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I like Serena said:Unfortunately, with $h(t)=c$, the function $H$ is actually concave up ($H''(x)\ge 0$), just notstrictlyconcave up.

But yes, it does illustrate that we cannot just make the function (strictly) concave up.

Perhaps there are differences in terminology...I was taught that a function is concave up on an interval when its second derivative is positive, and concave down when its second derivative is negative. When the derivative of a function is constant, then the function itself has no concavity.

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MermaidWonders

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Ah, so if we don't know if it's concave up or concave down, the statement is false?

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MermaidWonders said:Ah, so if we don't know if it's concave up or concave down, the statement is false?

Correct. (Nod)

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MermaidWonders said:

With the given information we can conclude that $H'(x)>0$ on the unit interval due to the

However, we have no information whatsoever on $H''$, meaning we cannot say anything about whether it's concave up or not.

So the statement could be true in some cases, but it is not true for the general case.

$h(t)=2-t^2$ is a counter example. We have $h(t)>0$ for $0\le t \le 1$, but $H''(x)=h'(x)=-2x < 0$ for $0<x\le 1$.

It means that the statement is false.

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MarkFL said:Perhaps there are differences in terminology...I was taught that a function is concave up on an interval when its second derivative is positive, and concave down when its second derivative is negative. When the derivative of a function is constant, then the function itself has no concavity.

I can only refer to the wiki page, which is consistent with what I've learned. And I've never heard or seen that it could be different. Can you perhaps provide a reference?

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MarkFL

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MermaidWonders said:

Let's look at an example that more clearly demonstrates the statement regarding concavity isn't always true. Consider:

\(\displaystyle H(x)=\ln(x+1)\)

And so:

\(\displaystyle H'(x)=h(x)=\frac{1}{x+1}\)

We have:

\(\displaystyle h(t)>0\) for $0\le t\le1$

But:

\(\displaystyle H''(x)=-(x+1)^{-2}\)

On the given interval, this function is concave down, but it meets the given criteria, so we know the statement is false.

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MarkFL

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I like Serena said:I can only refer to the wiki page, which is consistent with what I've learned. And I've never heard or seen that it could be different. Can you perhaps provide a reference?

I was simply taught that a linear function has no concavity...just recalling from memory. :)

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MarkFL said:I was simply taught that a linear function has no concavity...just recalling from memory. :)

I think we have

However, if we have $f''(x)=0$ only on part of the domain, the function can still be convex or concave.

For instance $f(x)=x^4$ counts as a

Moreover, it's even

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MermaidWonders

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Oh, I see. I finally get it now. Many thanks to all those who have helped me with this question! :)

Integral calculus is a branch of mathematics that deals with finding the area under a curve. It is used to solve problems involving continuous change, such as finding the velocity of an object or the volume of a 3D shape.

The difference between true and false integral calculus lies in the concept of integration. In true integral calculus, the limits of integration are fixed and definite, while in false integral calculus, the limits are variable and indefinite. This leads to different methods and results in solving problems.

To determine if an integral is true or false, you need to check the limits of integration. If they are fixed and definite, then it is a true integral. If the limits are variable and indefinite, then it is a false integral.

Integral calculus has various real-world applications, such as in physics, engineering, economics, and statistics. For example, it can be used to calculate the work done by a force or the total profit of a business. It is also used in optimization problems to find the maximum or minimum value of a function.

Some common techniques for solving true or false integral calculus problems include substitution, integration by parts, trigonometric substitution, and partial fractions. These techniques involve manipulating the integrand and using known integration rules to solve the integral.

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