If f(x)=5x^4-2x+7, then f'(x) equals:
- 20x^3-2 Correct answer.
- 5x^3-2
- 20x^3-2x
- 20x^4-2
Explanation
Differentiate term by term: d/dx[5x^4] = 20x^3, d/dx[-2x] = -2, and the constant 7 contributes 0. So f'(x) = 20x^3 - 2.
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Differentiate term by term: d/dx[5x^4] = 20x^3, d/dx[-2x] = -2, and the constant 7 contributes 0. So f'(x) = 20x^3 - 2.
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Chain rule for (g(x))^n: n(g(x))^{n-1} g'(x). Here g(x) = 3x + 1, so f'(x) = 5(3x + 1)^4 · 3 = 15(3x + 1)^4.
f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1) = 0 gives critical points x = ±1. Second derivative f''(x) = 6x: f''(-1) = -6 < 0 (local max), f''(1) = 6 > 0 (local min). So the local maximum is at x = -1.
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Velocity is the derivative of position: v(t) = 3t^2 - 12t + 9. Substituting t = 2: v(2) = 3·4 - 12·2 + 9 = 12 - 24 + 9 = -3.
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An antiderivative of 3x^2 is x^3. Evaluating the definite integral: [x^3]_0^2 = 8 - 0 = 8.
An antiderivative of cos x is sin x. Evaluating from 0 to π/2: sin(π/2) - sin(0) = 1 - 0 = 1.
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Compute ∫_0^4 x dx = [x^2/2]_0^4 = 8. Geometrically this is a right triangle with base 4 and height 4, so area = (1/2)·4·4 = 8.
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By the Fundamental Theorem of Calculus, ∫_a^b f(x) dx = F(b) - F(a) when F' = f. Here ∫_0^3 f = F(3) - F(0) = 8 - 2 = 6.
An antiderivative of 2x + 1 is x^2 + x. Evaluating from 0 to 1: (1 + 1) - (0 + 0) = 2.
Integration by parts: ∫ u dv = uv - ∫ v du. Set u = x, dv = e^x dx, so du = dx, v = e^x. Then ∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C = e^x(x - 1) + C.
Average value on [a, b] is (1/(b - a)) ∫_a^b f. Here (1/3) ∫_0^3 x^2 dx = (1/3)[x^3/3]_0^3 = (1/3)·9 = 3.
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Substitute x = 2: f(2) = 3(2)^2 - 5(2) + 2 = 3·4 - 10 + 2 = 12 - 10 + 2 = 4.
Factor x^2 - 6x + 5 = (x - 1)(x - 5). Setting each factor to zero gives roots x = 1 and x = 5. The quadratic formula with Δ = 36 - 20 = 16 yields the same roots: (6 ± 4)/2.
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Difference of squares: x^2 - 9 = (x - 3)(x + 3). Cancelling the common factor (x - 3) for x ≠ 3 leaves x + 3.
log_2(16) is the exponent that produces 16 when the base is 2. Since 2^4 = 16, log_2(16) = 4.
The exponential e^x is positive for every real x. As x → -∞, e^x → 0 but never reaches 0; as x → ∞, e^x → ∞. Therefore the range is the open interval (0, ∞).
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Discriminant Δ = b^2 - 4ac = (-4)^2 - 4(1)(5) = 16 - 20 = -4. A negative discriminant means no real roots.
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The exponential function is one-to-one, so e^{2x} = e^{x+3} implies 2x = x + 3. Subtracting x yields x = 3.
Arithmetic nth term: a_n = a_1 + (n - 1)d. Here a_5 = 3 + (5 - 1)·4 = 3 + 16 = 19.
Geometric nth term: a_n = a_1 r^{n-1}. Here a_4 = 2 · 3^{3} = 2 · 27 = 54.