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iren2701 [21]
3 years ago
6

Estimate and classify the critical points for the graph of each function.

Mathematics
1 answer:
Neko [114]3 years ago
6 0

Answer: B

(0.5, 7), maximum; (2, 1), point of inflection; (3.5, –5), minimum

Critical points of a function are those points at which a line drawn tangent to the curve is horizontal or vertical. On the graph, maximum and minimum points look like peaks and valleys, respectively. Points of inflection that are critical points are places where the curve flattens out and changes from being bent upward to being bent downward or vice versa.

Just passed the test

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A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is
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Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume V of a solid pyramid can be given as:

\displaystyle V = \frac{1}{3} \cdot b \cdot h,

where

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  • h is the height of the pyramid.

Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be x cm.

  • At the vertice of this pyramid, x = 0 and the length of a side of the square is also 0.
  • At the base of this pyramid, x = 3 and the length of a side of the square is 4 cm.

As a result, the length of a side of the square will be

\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x.

The area of the square will be

\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}.

Integrate the area of the horizontal cross-section with respect to x

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\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}.

In other words, the volume of this pyramid is 16 cubic centimeters.

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Step-by-step explanation:

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