Using derivatives, it is found that regarding the tangent line to the function, we have that:
- The equation of the line is y = 962x - 5119.
<h3>What is a linear function?</h3>
A linear function is modeled by:
y = mx + b
In which:
- m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
- b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.
The slope of the line tangent to a function f(x) at x = x' is given by f'(x'). In this problem, the function is given by:
f(x) = 5x³ + 2x + 1.
The derivative is given by:
f'(x) = 15x² + 2.
Hence the slope at x = 8 is:
m = f'(8) = 15(8)² + 2 = 962.
The line goes through the point (8,f(8)), hence:
f(8) = 5(8)³ + 2(8) + 1 = 2577.
Hence:
y = 962x + b
2577 = 962(8) + b
b = -5119.
Hence the equation is:
y = 962x - 5119.
More can be learned about tangent lines at brainly.com/question/8174665
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Answer:
272.973820315
Step-by-step explanation:
Answer:
2.25
Step-by-step explanation:
N=-5 you got to get the n by it self
Answer:
Step-by-step explanation:
The average scores s (on a 100 point scale) for the students can be modeled by
s = 75 - 6 In(t + 1), 0 < t < 12
where t is the time in months.
a) Since the students were given an exam and then retested monthly with equivalent exams, then,
For the original exam, t = 0
Therefore,
s = 75 - 6 In(0 + 1) = 75 - 6 In1
s = 75 - 6 × 0 = 75
b) the average score after 4 months, t = 4
Therefore,
s = 75 - 6 In(4 + 1) = 75 - 6 In5
s = 75 - 9.66 = 65.34
c) s = 60
Therefore,
60 = 75 - 6 In(t + 1)
6 In(t + 1) = 75 - 60 = 15
In(t + 1) = 15/6 = 2.5
t + 1 = e^2.5 = 12.18
t = 12.18 - 1 = 11.18
t = 11 approximately
