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3 years ago

Find the derivative of f(x) = -2x2 + 11x at x = 9.

Mathematics

1 answer:

Marianna [84]3 years ago

8 0

Answer:

The value of the derivative at x = 9 is -25

Step-by-step explanation:

Here in this question, we are told to find the derivative of the given equation at the point where x = 9

What we need to do here is simply, differentiate f(x), then substitute that value x = 9 into the differentiated equation

In notation form, what we are to find is f’(9)

So;

f(x) = -2x^2 + 11x

f’(x) = -4x + 11

So therefore;

f’(9) = -4(9) + 11

f’(9) = -36 + 11

f’(9) = 11-36

f’(9) = -25

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Answer:

She earned $60

Step-by-step explanation:

Since she had 20 dollars remaining from may, she earned the rest of the money in june so:

80-20=60

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3 years ago

Use CALCULUS to find coordinates of the turning point on C.

antiseptic1488 [7]

Oi

Just calculating the differential. Step by step:

$y=12 \sqrt{x} -x \frac{3}{2}-10 \\ \\ y=12x^{ \frac{1}{2} } -x \frac{3}{2}-10 \ \ \ \boxed{transform \ \sqrt{x} =x^{ \frac{1}{2} }} \\ \\ Differentiating \\ \\ y'=12. \frac{1}{2}.x^{ \frac{1}{2}-1 } -1.x^{1-1}. \frac{3}{2}-0 \\ \\ y'=6.x^{ -\frac{1}{2} } -x^{0}. \frac{3}{2} \\ \\ y'=6. \frac{1}{x^{ \frac{1}{2} }} -1. \frac{3}{2} \\ \\ \boxed{y'=\frac{6}{ \sqrt{x} } -\frac{3}{2}}$

If you want to know where the tangent has no slope:

$\frac{6}{ \sqrt{x} } - \frac{3}{2} =0 \\ \\ \frac{6}{ \sqrt{x} } = \frac{3}{2} \\ \\ 3 \sqrt{x} =12 \\ \\ \sqrt{x} = \frac{12}{3} \\ \\ \sqrt{x} =4 \\ \\ (\sqrt{x})^2 =4^2 \\ \\ x=16$

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3 years ago

63,825 in scientific notation

valentinak56 [21]

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3 years ago

If f(x) = x2 -1, what is the equation for f–1(x)?

professor190 [17]

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3 years ago

[PLEASE ANSWER BOTH QUESTIONS] The following data points represent the number of quesadillas each person at Toby's Tacos ate. So

Fynjy0 [20]

Answer:

Part 1: Already solved.

Part 2: 1 quesadilla.

Step-by-step explanation:

Part 1: The data points given are 0, 0, 1/4, 1/2, 1/2, 1, 1, 1, 5/4, 2, and 2.

The data are already sorted from least to greatest.

Part 2: To get the interquartile range, we must know the first and third quartiles. To find those, we need to know the median of the data.

There are 11 data points, so we can cross out 5 numbers on the left and cross out 5 numbers on the right. That leaves a median of 1.

Since we know the median is 1, we can find the first quartile by finding the median of the following numbers: 0, 0, 1/4, 1/2, 1/2 (any number less than 1). There are five numbers, so cross out two numbers from the left and cross out two numbers from the right. We are left with a first quartile of 1/4.

Now that we have the first quartile, we need to find the third quartile. Since the median was actually the first "1" in the data set, the third quartile will be the median of the numbers to the right of that "1": 1, 1, 5/4, 2, 2. Cross out the leftmost two numbers, and cross out the rightmost two numbers. We are left with a third quartile of 5/4.

Now that we have both the third and first quartiles, we can find the interquartile range! The IQR is calculated by finding the third quartile minus the first quartile. 5/4 - 1/4 = 4/4 = 1.

So, your interquartile range is 1 quesadilla.

Hope this helps!

5 0

3 years ago

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