Calculus BC · Unit 2
Match a limit to f′(a), or evaluate the limit by recognizing the derivative.
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Two forms of the derivative
\(f'(a) = \lim_{h \to 0} \dfrac{f(a + h) - f(a)}{h}\)
\(f'(a) = \lim_{x \to a} \dfrac{f(x) - f(a)}{x - a}\)
Match the point and the function
The number subtracted in the numerator is \(f(a)\). The denominator must approach 0. A limit that uses \(f(h)\) in place of \(f(a)\), or a denominator that does not approach 0, is not \(f'(a)\).
A limit can be a derivative you already know
If the limit is the definition of \(f'(a)\), you may evaluate \(f'(a)\) with a derivative rule instead of expanding.
The value of the derivative is not the function value
\(f'(a) = 4\) does not say \(f(a) = 4\).
Pick the limit that is exactly f′(a).
The h-form of the derivative at 4 is
\(f'(4) = \lim_{h \to 0} \dfrac{f(4 + h) - f(4)}{h}\)
The other choices change the function value, the sign, or the place where h approaches.
Answer: B
\(f'(-1) = \lim_{h \to 0} \dfrac{f(-1 + h) - f(-1)}{h}\)
Answer: A
The x-form at 5 is
\(g'(5) = \lim_{x \to 5} \dfrac{g(x) - g(5)}{x - 5}\)
Choice D is the negative of that limit.
Answer: A
The h-form uses \(f(2 + h) - f(2)\).
\(f(x) = x^3\), because \(f(2) = 8\).
\(f'(x) = 3x^2\)
\(f'(2) = 3 \cdot 4 = 12\)
Answer: \(f(x) = x^3\), and the limit is \(12\)
This is the x-form of \(f'(4)\) for \(f(x) = \sqrt{x}\).
\(f(4) = 2\)
\(f'(x) = \dfrac{1}{2\sqrt{x}}\)
\(f'(4) = \dfrac{1}{2 \cdot 2} = \dfrac{1}{4}\)
Answer: \(\dfrac{1}{4}\)
The limit is the definition of \(f'(3)\).
\(f'(3) = -7\)
Answer: \(-7\)
Evaluate the limit by naming the derivative.
This is \(f'\bigl(\tfrac{\pi}{3}\bigr)\) for \(f(x) = \sin x\).
\(f'(x) = \cos x\)
\(\cos\!\bigl(\tfrac{\pi}{3}\bigr) = \dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
This is \(f'(2)\) for \(f(x) = e^x\).
\(f'(x) = e^x\)
\(f'(2) = e^2\)
Answer: \(e^2\)
This is \(f'(5)\) for \(f(x) = \ln x\).
\(f'(x) = \dfrac{1}{x}\)
\(f'(5) = \dfrac{1}{5}\)
Answer: \(\dfrac{1}{5}\)
\(x + 2 = x - (-2)\)
This is \(f'(-2)\) for \(f(x) = x^2\), since \(f(-2) = 4\).
\(f'(x) = 2x\)
\(f'(-2) = -4\)
Answer: \(-4\)
This is \(f'(3)\) for \(f(x) = x^2\).
\(f'(x) = 2x\)
\(f'(3) = 6\)
Answer: \(6\)
The limit is the definition of \(f'(1)\).
It does not give \(f(1)\), an average rate on a fixed interval, or the limit of f.
Answer: B. \(f'(1) = 4\)
A disguised derivative, or a limit that is almost the definition.
A is the h-form at 3.
B is the x-form at 3.
D uses x as the dummy step, so it is the same as the h-form.
C has denominator \(h + 3\), which approaches 3, not 0.
C is not \(f'(3)\).
Answer: C
This is \(f'(4)\) for \(f(x) = \sqrt{x}\).
\(f'(x) = \dfrac{1}{2\sqrt{x}}\)
\(f'(4) = \dfrac{1}{4}\)
Answer: \(\dfrac{1}{4}\)
This is \(f'(1)\) for \(f(x) = x^2 + 4x\), because \(f(1) = 5\).
\(f'(x) = 2x + 4\)
\(f'(1) = 6\)
Check by factoring:
\(x^2 + 4x - 5 = (x + 5)(x - 1)\)
\(x + 5\) at \(x = 1\) is 6.
Answer: \(6\)
This is \(f'(1)\) for \(f(x) = \dfrac{1}{x}\).
\(f'(x) = -\dfrac{1}{x^2}\)
\(f'(1) = -1\)
Algebra check:
\(\dfrac{1}{1 + h} - 1 = \dfrac{1 - (1 + h)}{1 + h} = \dfrac{-h}{1 + h}\)
\(\dfrac{-h}{h(1 + h)} = -\dfrac{1}{1 + h}\) for \(h \neq 0\)
The limit as \(h \to 0\) is \(-1\).
Answer: \(-1\)
Differentiability at a single point implies continuity at that point.
The derivative does not have to be 0.
Nothing is claimed about other x-values.
The derivative itself does not have to be continuous.
Answer: A. f is continuous at \(x = a\)
This is \(f'(\pi)\) for \(f(x) = \cos x\).
\(f'(x) = -\sin x\)
\(-\sin \pi = 0\)
Answer: \(0\)
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