Calculus BC · Unit 2
The slope is f′(a) and the point is (a, f(a)). A horizontal tangent is the line y = f(a) where f′(a) = 0.
Point-slope form
\(y - y_1 = m(x - x_1)\)
The point is \(\bigl(a, f(a)\bigr)\) and the slope is \(f'(a)\).
Slope before the line
Write \(f'(x)\). Substitute a. Then write \(f(a)\) on its own line. Then write the line.
A horizontal tangent has slope 0
Solve \(f'(x) = 0\). Each solution x = a gives a horizontal line \(y = f(a)\).
Parallel means the same slope
A tangent parallel to \(y = mx + b\) has slope m. Set \(f'(x)\) equal to m, then use the point on the curve.
One point. Power rule or a basic derivative.
\(y - y_1 = m(x - x_1)\)
\(f'(x) = 2x\)
\(m = f'(3) = 6\)
\(f(3) = 9\)
\(y - 9 = 6(x - 3)\)
\(y = 6x - 9\)
Answer: \(y - 9 = 6(x - 3)\)
\(f'(x) = 3x^2 - 6\)
\(f'(2) = 12 - 6 = 6\)
\(f(2) = 8 - 12 = -4\)
\(y - (-4) = 6(x - 2)\)
\(y + 4 = 6(x - 2)\)
Answer: \(y + 4 = 6(x - 2)\)
\(f'(x) = \dfrac{1}{2\sqrt{x}}\)
\(f'(4) = \dfrac{1}{4}\)
\(f(4) = 2\)
\(y - 2 = \dfrac{1}{4}(x - 4)\)
Answer: \(y - 2 = \dfrac{1}{4}(x - 4)\)
\(f'(x) = e^x\)
\(f'(0) = 1\)
\(f(0) = 1\)
\(y - 1 = 1(x - 0)\)
\(y = x + 1\)
Answer: \(y = x + 1\)
\(f'(x) = \dfrac{1}{x}\)
\(f'(e) = \dfrac{1}{e}\)
\(f(e) = 1\)
\(y - 1 = \dfrac{1}{e}(x - e)\)
Answer: \(y - 1 = \dfrac{1}{e}(x - e)\)
\(f'(x) = \cos x\)
\(f'(0) = 1\)
\(f(0) = 0\)
\(y - 0 = 1(x - 0)\)
\(y = x\)
Answer: \(y = x\)
A horizontal tangent, or a product evaluated at one point.
\(f'(x) = 3x^2\)
\(f'(1) = 3\)
\(f(1) = 1\)
\(y - 1 = 3(x - 1)\)
\(y = 3x - 2\)
Answer: A. \(y = 3x - 2\)
\(f'(x) = 3x^2 - 3\)
\(3(x^2 - 1) = 0\)
\(x = 1\) or \(x = -1\)
\(f(1) = 1 - 3 = -2\)
\(f(-1) = -1 + 3 = 2\)
The horizontal lines are \(y = -2\) and \(y = 2\).
Answer: \(y = -2\) and \(y = 2\)
\(f'(x) = -\dfrac{1}{x^2}\)
\(f'(2) = -\dfrac{1}{4}\)
\(f(2) = \dfrac{1}{2}\)
\(y - \dfrac{1}{2} = -\dfrac{1}{4}(x - 2)\)
Answer: \(y - \dfrac{1}{2} = -\dfrac{1}{4}(x - 2)\)
\(f'(x) = 2x - 6\)
\(2x - 6 = 0\)
\(x = 3\)
\(f(3) = 9 - 18 + 1 = -8\)
\(y = -8\)
Answer: \(x = 3\), \(y = -8\)
\(f'(x) = \cos x - x\sin x\)
\(f'(\pi) = \cos\pi - \pi\sin\pi = -1\)
\(f(\pi) = \pi\cos\pi = -\pi\)
\(y - (-\pi) = -1(x - \pi)\)
\(y + \pi = -(x - \pi)\)
Answer: \(y + \pi = -(x - \pi)\)
\(f'(x) = \dfrac{1}{x}\)
\(f'(1) = 1\)
\(f(1) = 0\)
\(y - 0 = 1(x - 1)\)
\(y = x - 1\)
Answer: \(y = x - 1\)
Match a slope, or use a product at a special point.
\((uv)' = u'v + uv'\)
\(f'(x) = e^x\sin x + e^x\cos x = e^x(\sin x + \cos x)\)
\(f'(0) = e^0(0 + 1) = 1\)
\(f(0) = e^0\sin 0 = 0\)
\(y - 0 = 1(x - 0)\)
\(y = x\)
Answer: \(y = x\)
Parallel lines have the same slope, so \(f'(x) = 8\).
\(f'(x) = 2x\)
\(2x = 8\)
\(x = 4\)
\(f(4) = 16\)
\(y - 16 = 8(x - 4)\)
\(y = 8x - 16\)
Answer: point \((4, 16)\), line \(y = 8x - 16\)
\(f'(x) = 2x\)
\(f'(2) = 4\)
The given line already has slope 4, so the slopes match at \(x = 2\).
The point \((2, f(2)) = (2, 4)\) must lie on the line.
\(4 = 4\cdot 2 + b\)
\(b = -4\)
Answer: \(b = -4\)
\(f'(x) = -\sin x\)
\(f'\!\bigl(\tfrac{\pi}{3}\bigr) = -\dfrac{\sqrt{3}}{2}\)
\(f\!\bigl(\tfrac{\pi}{3}\bigr) = \dfrac{1}{2}\)
\(y - \dfrac{1}{2} = -\dfrac{\sqrt{3}}{2}\left(x - \dfrac{\pi}{3}\right)\)
Answer: \(y - \dfrac{1}{2} = -\dfrac{\sqrt{3}}{2}\left(x - \dfrac{\pi}{3}\right)\)
\(f'(x) = \dfrac{(x + 1) - x}{(x + 1)^2} = \dfrac{1}{(x + 1)^2}\)
\(f'(1) = \dfrac{1}{4}\)
\(f(1) = \dfrac{1}{2}\)
\(y - \dfrac{1}{2} = \dfrac{1}{4}(x - 1)\)
Answer: \(y - \dfrac{1}{2} = \dfrac{1}{4}(x - 1)\)
\(f'(x) = 3x^2 + 1\)
\(f'(-1) = 3 + 1 = 4\)
\(f(-1) = -1 - 1 = -2\)
\(y - (-2) = 4\bigl(x - (-1)\bigr)\)
\(y + 2 = 4(x + 1)\)
\(y = 4x + 2\)
Answer: A. \(y = 4x + 2\)
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