1. Use the Mean Value Theorem to find an x-value where the instantaneous slope=average slope over the given interval:
f(x)=(9-x^2)^(1/2) on [-3,2]
I have done three chapters of Calculus already and I never have had to check for extraneous roots. So, my question is, when exactly do I check for extraneous roots? (ex/ when I was looking for critical points and I root my answer, I didn't have to check for extraneous roots)
2. Can someone show me how e^(ln((e^5-1)/5))=(e^5-1)/5?
by definition, e^(ln x) = x
for the MVT,
f(-3) = 0
f(2) = √5
average rate is thus √5/5 or 1/√5
So, you want to find where the slope of f is 1/√5
f' = -x/√(9-x^2)
so, find x where f' = 1/√5
-x/√(9-x^2) = 1/√5
square both sides. (This is where an extraneous root might appear, because both (√a)^2 = a and (-√a)^2 = a)
x^2/(9-x^2) = 1/5
5x^2 = 9-x^2
6x^2 = 9
x = ±3/√6
Now you have to check which solution you want.
f'(3/√6) = -(3/√6)/√(9 - 9/6)
= -3/√6 / √(45/6)
= -3/√45
= -1/√5
And of course, f'(-3/√6) = +1/√5
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%289-x^2%29%2C+y%3D1%2F%E2%88%9A5%28x%2B3%29%2C+y+%3D+1%2F%E2%88%9A5+%28x+%2B+3%2F%E2%88%9A6%29+%2B+%E2%88%9A%2845%2F6%29+where+-3+%3C%3Dx+%3C%3D+2
1. When to check for extraneous roots is not directly related to finding critical points, but rather to solving equations. In general, you need to check for extraneous roots when solving equations that involve taking roots or other operations that might introduce additional solutions. This is because some of these operations can generate solutions that are not valid in the original context.
In the case of finding critical points, you are looking for points where the derivative is either zero or undefined. When you solve for the derivative and set it equal to zero, you are essentially solving an equation. At this point, you don't need to check for extraneous roots because the equation was derived directly from the original function and its derivative.
However, if you were to solve for x in a different equation that involves taking a root, for example, you might need to check for extraneous roots. This is because taking a root can introduce multiple solutions, and not all of them may satisfy the original equation.
In general, it's a good practice to always check the obtained solutions by substituting them back into the original equation to make sure they are valid.
2. To show that e^(ln((e^5-1)/5)) is equal to (e^5-1)/5, we can simplify the expression using the properties of logarithms and exponents.
Step 1: Start with the expression e^(ln((e^5-1)/5)).
Step 2: Apply the property of logarithms, ln(a^b) = b * ln(a), to simplify the expression inside the logarithm: ln((e^5-1)/5) = ln(e^5-1) - ln(5).
Step 3: Apply the property of exponents, e^(ln(a)) = a, to the expression inside the exponential: e^(ln(e^5-1) - ln(5)).
Step 4: Apply the property of exponents, e^(a-b) = e^a / e^b, to separate the terms inside the exponential: e^(ln(e^5-1)) / e^(ln(5)).
Step 5: Apply the property of logarithms, ln(e^x) = x, and the definition of the natural logarithm, ln(e) = 1, to simplify the expression: (e^5-1) / 5.
Therefore, e^(ln((e^5-1)/5)) is equal to (e^5-1)/5.