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  1. AP Calculus
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What does the Inverse Function Theorem state?

If fff is differentiable at aaa and f′(a)≠0f'(a) \neq 0f′(a)=0, then f−1f^{-1}f−1 is differentiable at f(a)f(a)f(a) and (f−1)′(f(a))=1f′(a)(f^{-1})'(f(a)) = \frac{1}{f'(a)}(f−1)′(f(a))=f′(a)1​.

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What does the Inverse Function Theorem state?

If fff is differentiable at aaa and f′(a)≠0f'(a) \neq 0f′(a)=0, then f−1f^{-1}f−1 is differentiable at f(a)f(a)f(a) and (f−1)′(f(a))=1f′(a)(f^{-1})'(f(a)) = \frac{1}{f'(a)}(f−1)′(f(a))=f′(a)1​.

How does the Intermediate Value Theorem relate to inverse functions?

If fff is continuous on [a,b][a, b][a,b] and kkk is any number between f(a)f(a)f(a) and f(b)f(b)f(b), then there exists at least one ccc in [a,b][a, b][a,b] such that f(c)=kf(c) = kf(c)=k. This helps establish the existence of an inverse function over an interval.

What does the Mean Value Theorem state?

If fff is continuous on [a,b][a, b][a,b] and differentiable on (a,b)(a, b)(a,b), then there exists a ccc in (a,b)(a, b)(a,b) such that f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}f′(c)=b−af(b)−f(a)​.

What does the chain rule state?

If y=f(u)y = f(u)y=f(u) and u=g(x)u = g(x)u=g(x) are both differentiable, then dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

What does the quotient rule state?

If h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}h(x)=g(x)f(x)​, then h′(x)=g(x)f′(x)−f(x)g′(x)[g(x)]2h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}h′(x)=[g(x)]2g(x)f′(x)−f(x)g′(x)​.

What does the power rule state?

If f(x)=xnf(x) = x^nf(x)=xn, then f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1.

What does the constant multiple rule state?

If f(x)=c⋅g(x)f(x) = c \cdot g(x)f(x)=c⋅g(x), where ccc is a constant, then f′(x)=c⋅g′(x)f'(x) = c \cdot g'(x)f′(x)=c⋅g′(x).

What does the sum/difference rule state?

If h(x)=f(x)±g(x)h(x) = f(x) \pm g(x)h(x)=f(x)±g(x), then h′(x)=f′(x)±g′(x)h'(x) = f'(x) \pm g'(x)h′(x)=f′(x)±g′(x).

What does the product rule state?

If h(x)=f(x)g(x)h(x) = f(x)g(x)h(x)=f(x)g(x), then h′(x)=f′(x)g(x)+f(x)g′(x)h'(x) = f'(x)g(x) + f(x)g'(x)h′(x)=f′(x)g(x)+f(x)g′(x).

What is the difference between finding f′(x)f'(x)f′(x) and (f−1)′(x)(f^{-1})'(x)(f−1)′(x)?

f′(x)f'(x)f′(x) is the derivative of the original function. (f−1)′(x)(f^{-1})'(x)(f−1)′(x) requires using the inverse derivative formula: (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}(f−1)′(x)=f′(f−1(x))1​.

Compare the chain rule and the inverse function derivative rule.

Chain rule: used for composite functions. Inverse function rule: specifically for derivatives of inverse functions.

What is the difference between implicit and explicit differentiation?

Explicit: function is defined as y = f(x). Implicit: function is not explicitly solved for y.

Compare finding the derivative of f(x)f(x)f(x) and finding the derivative of f−1(x)f^{-1}(x)f−1(x) when f(x)f(x)f(x) is given explicitly.

For f(x)f(x)f(x), directly apply differentiation rules. For f−1(x)f^{-1}(x)f−1(x), use the inverse derivative formula or find f−1(x)f^{-1}(x)f−1(x) explicitly and then differentiate.

Compare finding the derivative of f(x)f(x)f(x) and finding the derivative of f−1(x)f^{-1}(x)f−1(x) when f(x)f(x)f(x) is given implicitly.

For f(x)f(x)f(x), use implicit differentiation directly. For f−1(x)f^{-1}(x)f−1(x), either find f−1(x)f^{-1}(x)f−1(x) explicitly (if possible) and then differentiate, or use the inverse derivative formula in conjunction with implicit differentiation.

What is the difference between finding f−1(x)f^{-1}(x)f−1(x) and finding (f−1)′(x)(f^{-1})'(x)(f−1)′(x)?

Finding f−1(x)f^{-1}(x)f−1(x) involves switching xxx and yyy and solving for yyy. Finding (f−1)′(x)(f^{-1})'(x)(f−1)′(x) involves differentiating the inverse function using the inverse derivative rule.

Compare the derivatives of f(x)f(x)f(x) and f−1(x)f^{-1}(x)f−1(x) at corresponding points.

If f(a)=bf(a) = bf(a)=b, then f′(a)f'(a)f′(a) is the slope of f(x)f(x)f(x) at x=ax=ax=a, and (f−1)′(b)(f^{-1})'(b)(f−1)′(b) is the slope of f−1(x)f^{-1}(x)f−1(x) at x=bx=bx=b. These slopes are reciprocals of each other.

What is the difference between the graph of f(x)f(x)f(x) and the graph of f′(x)f'(x)f′(x)?

The graph of f(x)f(x)f(x) shows the function's values, while the graph of f′(x)f'(x)f′(x) shows the rate of change of f(x)f(x)f(x).

Compare the domain and range of a function and its inverse.

The domain of f(x)f(x)f(x) is the range of f−1(x)f^{-1}(x)f−1(x), and the range of f(x)f(x)f(x) is the domain of f−1(x)f^{-1}(x)f−1(x).

What is the difference between finding f′(a)f'(a)f′(a) and (f−1)′(a)(f^{-1})'(a)(f−1)′(a)?

f′(a)f'(a)f′(a) is the derivative of f(x)f(x)f(x) evaluated at x=ax=ax=a. (f−1)′(a)(f^{-1})'(a)(f−1)′(a) is the derivative of the inverse function evaluated at x=ax=ax=a, and it requires using the inverse derivative formula.

What is an inverse function?

A function that 'reverses' another function. If f(a)=bf(a) = bf(a)=b, then f−1(b)=af^{-1}(b) = af−1(b)=a.

What does it mean for a function to be differentiable?

A function is differentiable at a point if its derivative exists at that point.

What is an invertible function?

A function that has an inverse function.

Define the derivative of a function.

The derivative of a function f(x)f(x)f(x) is a measure of how f(x)f(x)f(x) changes as xxx changes.

What is a tangent line?

A line that touches a curve at a point and has the same slope as the curve at that point.

What is the point-slope form of a line?

The equation of a line given a point (x1,y1)(x_1, y_1)(x1​,y1​) and slope mmm: y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​).

What is the domain of a function?

The set of all possible input values (x-values) for which the function is defined.

What is the range of a function?

The set of all possible output values (y-values) of the function.

What does strictly increasing mean?

A function f(x)f(x)f(x) is strictly increasing if, for any x1<x2x_1 < x_2x1​<x2​, we have f(x1)<f(x2)f(x_1) < f(x_2)f(x1​)<f(x2​).

What is the reciprocal of a number?

The reciprocal of a number xxx is 1/x1/x1/x.