Calculus 1 Quiz: Power Rule
12 questions · exam conditions
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Power RuleQuestion 1 of 12

The line tangent to the curve y=x2+cy = x^2 + c at x=2x=2 passes through the point (1,1)(1, 1). What is the value of cc?

00
−3-3
−1-1
11
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Calculus 1 Quiz

Calculus 1 Quiz: Power Rule

Practice Power Rule in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Power Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The line tangent to the curve y=x2+cy = x^2 + c at x=2x=2 passes through the point (1,1)(1, 1). What is the value of cc?

  1. 00
  2. −3-3
  3. −1-1
  4. 11 (correct answer)
Explanation: This problem combines tangent line equations with point conditions, testing your ability to use both the derivative and the point-slope form of a line. To find the tangent line at x=2x = 2, you need both a point on the curve and the slope at that point. The derivative y′=2xy' = 2x gives the slope, so at x=2x = 2, the slope is 2(2)=42(2) = 4. The point on the curve is (2,4+c)(2, 4 + c) since y=22+c=4+cy = 2^2 + c = 4 + c. Using point-slope form with slope m=4m = 4 and point (2,4+c)(2, 4 + c): y−(4+c)=4(x−2)y - (4 + c) = 4(x - 2) y=4x−8+4+c=4x−4+cy = 4x - 8 + 4 + c = 4x - 4 + c Since this tangent line passes through (1,1)(1, 1), substitute these coordinates: 1=4(1)−4+c1 = 4(1) - 4 + c 1=4−4+c1 = 4 - 4 + c 1=c1 = c Therefore, c=1c = 1, which is choice D. Choice A (c=0c = 0) would give tangent line y=4x−4y = 4x - 4, which equals 0 at x=1x = 1, not 1. Choice B (c=−3c = -3) gives y=4x−7y = 4x - 7, which equals −3-3 at x=1x = 1. Choice C (c=−1c = -1) gives y=4x−5y = 4x - 5, which equals −1-1 at x=1x = 1. Strategy tip: When a tangent line must pass through a given point, always set up the tangent line equation first using the derivative for slope, then substitute the given point's coordinates to solve for unknown parameters.

Question 2

Let f(x)=axn+bxn−1f(x) = ax^n + bx^{n-1} where aa and bb are non-zero real numbers and nn is an integer greater than 2. What is the smallest integer kk for which the kk-th derivative, f(k)(x)f^{(k)}(x), is a non-zero constant?

  1. n−1n-1
  2. nn (correct answer)
  3. n+1n+1
  4. n−2n-2
Explanation: Consider the term axnax^n. Each time we differentiate, the power of xx decreases by 1. After differentiating nn times, the term becomes a⋅n!⋅x0=a⋅n!a \cdot n! \cdot x^0 = a \cdot n!, which is a non-zero constant. Now consider the term bxn−1bx^{n-1}. After differentiating n−1n-1 times, it becomes b⋅(n−1)!b \cdot (n-1)!, which is a non-zero constant. After differentiating nn times, this term becomes 0. Therefore, the nn-th derivative of the entire function is f(n)(x)=a⋅n!+0=a⋅n!f^{(n)}(x) = a \cdot n! + 0 = a \cdot n!. This is a non-zero constant. For any k<nk < n, at least the first term will still contain a power of xx, so f(k)(x)f^{(k)}(x) will not be a constant. For k=n+1k=n+1, the derivative will be 0. Thus, the smallest integer kk for which the derivative is a non-zero constant is k=nk=n.

Question 3

Let f(x)=xaxbf(x) = \frac{x^a}{x^b} where a=3/2a = 3/2 and b=−1/2b = -1/2. If g(x)g(x) is the derivative of f(x)f(x), what is g(16)g(16)?

  1. 3232 (correct answer)
  2. 88
  3. 11
  4. 332\frac{3}{32}
Explanation: First, simplify the function f(x)f(x) using the rules of exponents: f(x)=xa−bf(x) = x^{a-b}. Substitute the given values of aa and bb: a−b=32−(−12)=32+12=42=2a-b = \frac{3}{2} - (-\frac{1}{2}) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2. So, the function simplifies to f(x)=x2f(x) = x^2. The function g(x)g(x) is the derivative of f(x)f(x). Using the Power Rule, g(x)=f′(x)=2xg(x) = f'(x) = 2x. Finally, evaluate g(x)g(x) at x=16x=16: g(16)=2(16)=32g(16) = 2(16) = 32.

Question 4

Let f(x)=x3−xxf(x) = \frac{x^3 - \sqrt{x}}{x}. What is the value of f′(4)f'(4)?

  1. 314\frac{31}{4}
  2. 12716\frac{127}{16}
  3. 12916\frac{129}{16} (correct answer)
  4. 88
Explanation: First, rewrite the function f(x)f(x) by splitting the fraction: f(x)=x3x−xx=x2−x1/2x−1=x2−x−1/2f(x) = \frac{x^3}{x} - \frac{\sqrt{x}}{x} = x^2 - x^{1/2}x^{-1} = x^2 - x^{-1/2}. Now, apply the Power Rule to find the derivative: f′(x)=2x−(−12)x−3/2=2x+12x−3/2f'(x) = 2x - (-\frac{1}{2})x^{-3/2} = 2x + \frac{1}{2}x^{-3/2}. Finally, evaluate the derivative at x=4x=4: f′(4)=2(4)+12(4)−3/2=8+12⋅1(41/2)3=8+12⋅123=8+12⋅18=8+116=12816+116=12916f'(4) = 2(4) + \frac{1}{2}(4)^{-3/2} = 8 + \frac{1}{2} \cdot \frac{1}{(4^{1/2})^3} = 8 + \frac{1}{2} \cdot \frac{1}{2^3} = 8 + \frac{1}{2} \cdot \frac{1}{8} = 8 + \frac{1}{16} = \frac{128}{16} + \frac{1}{16} = \frac{129}{16}.

Question 5

The curve given by the equation y=13x3−32x2−4x+1y = \frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x + 1 has horizontal tangent lines at two distinct points. What is the sum of the x-coordinates of these points?

  1. −3-3
  2. 33 (correct answer)
  3. −4-4
  4. 44
Explanation: A horizontal tangent line occurs where the derivative is equal to zero. First, find the derivative of the function using the Power Rule: y′=ddx(13x3−32x2−4x+1)=x2−3x−4y' = \frac{d}{dx}(\frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x + 1) = x^2 - 3x - 4. Set the derivative equal to zero to find the x-coordinates: x2−3x−4=0x^2 - 3x - 4 = 0. This quadratic equation can be factored as (x−4)(x+1)=0(x-4)(x+1) = 0, which gives solutions x=4x=4 and x=−1x=-1. The sum of these x-coordinates is 4+(−1)=34 + (-1) = 3.

Question 6

Let g(x)=2x3+3x2g(x) = \frac{2}{x^3} + 3x^2. Find the value of the second derivative, g′′(1)g''(1).

  1. 3030 (correct answer)
  2. 1818
  3. 00
  4. −18-18
Explanation: First, rewrite the function in terms of powers: g(x)=2x−3+3x2g(x) = 2x^{-3} + 3x^2. Now, find the first derivative using the Power Rule: g′(x)=2(−3)x−4+3(2)x1=−6x−4+6xg'(x) = 2(-3)x^{-4} + 3(2)x^1 = -6x^{-4} + 6x. Next, find the second derivative by differentiating g′(x)g'(x): g′′(x)=−6(−4)x−5+6=24x−5+6g''(x) = -6(-4)x^{-5} + 6 = 24x^{-5} + 6. Finally, evaluate g′′(x)g''(x) at x=1x=1: g′′(1)=24(1)−5+6=24+6=30g''(1) = 24(1)^{-5} + 6 = 24 + 6 = 30.

Question 7

Let h(x)=(x2+1)(1x+x)h(x) = (x^2+1)(\frac{1}{x} + x). Find the slope of the tangent line to the graph of h(x)h(x) at x=−1x=-1.

  1. 44 (correct answer)
  2. 66
  3. 00
  4. −4-4
Explanation: To find the slope of the tangent line, we need to find the derivative h′(x)h'(x). It is easiest to first expand and simplify the expression for h(x)h(x): h(x)=x2(1x)+x2(x)+1(1x)+1(x)=x+x3+x−1+x=x3+2x+x−1h(x) = x^2(\frac{1}{x}) + x^2(x) + 1(\frac{1}{x}) + 1(x) = x + x^3 + x^{-1} + x = x^3 + 2x + x^{-1}. Now, apply the Power Rule to find the derivative: h′(x)=3x2+2−1x−2=3x2+2−1x2h'(x) = 3x^2 + 2 - 1x^{-2} = 3x^2 + 2 - \frac{1}{x^2}. Finally, evaluate the derivative at x=−1x=-1 to find the slope: h′(−1)=3(−1)2+2−1(−1)2=3(1)+2−11=3+2−1=4h'(-1) = 3(-1)^2 + 2 - \frac{1}{(-1)^2} = 3(1) + 2 - \frac{1}{1} = 3 + 2 - 1 = 4.

Question 8

Let f(x)=xkf(x) = x^k and g(x)=xk+2g(x) = x^{k+2}, where kk is a positive integer. If f′(2)=g′(1)f'(2) = g'(1), what is the value of kk?

  1. 11
  2. 44
  3. 33
  4. 22 (correct answer)
Explanation: This problem tests your ability to find derivatives using the power rule and solve equations involving derivative values. When you see functions with exponents and need to compare their derivatives at specific points, you'll want to differentiate first, then substitute the given values. Let's find the derivatives of both functions. Using the power rule, f′(x)=kxk−1f'(x) = kx^{k-1} and g′(x)=(k+2)xk+1g'(x) = (k+2)x^{k+1}. Now we can evaluate each derivative at the specified points: f′(2)=k⋅2k−1f'(2) = k \cdot 2^{k-1} and g′(1)=(k+2)⋅1k+1=k+2g'(1) = (k+2) \cdot 1^{k+1} = k+2. Since f′(2)=g′(1)f'(2) = g'(1), we have the equation k⋅2k−1=k+2k \cdot 2^{k-1} = k+2. Let's test the answer choices systematically. For choice A (k=1k=1): 1⋅20=1⋅1=11 \cdot 2^0 = 1 \cdot 1 = 1, but k+2=3k+2 = 3. Since 1≠31 \neq 3, this is incorrect. For choice B (k=4k=4): 4⋅23=4⋅8=324 \cdot 2^3 = 4 \cdot 8 = 32, but k+2=6k+2 = 6. Since 32≠632 \neq 6, this is incorrect. For choice C (k=3k=3): 3⋅22=3⋅4=123 \cdot 2^2 = 3 \cdot 4 = 12, but k+2=5k+2 = 5. Since 12≠512 \neq 5, this is incorrect. For choice D (k=2k=2): 2⋅21=2⋅2=42 \cdot 2^1 = 2 \cdot 2 = 4, and k+2=4k+2 = 4. Since 4=44 = 4, this works! When solving derivative problems involving equations, always substitute your answer back into the original condition to verify it satisfies the given relationship. This double-checking catches calculation errors and confirms your solution.

Question 9

The position of a particle moving along the x-axis is given by p(t)=t3−3t2−9t+1p(t) = t^3 - 3t^2 - 9t + 1. For what positive value of tt is the particle's acceleration equal to zero?

  1. 33
  2. 22
  3. 11 (correct answer)
  4. 12\frac{1}{2}
Explanation: The velocity of the particle is the first derivative of the position function, and the acceleration is the second derivative. First, find the velocity function v(t)=p′(t)v(t) = p'(t): v(t)=3t2−6t−9v(t) = 3t^2 - 6t - 9. Next, find the acceleration function a(t)=v′(t)=p′′(t)a(t) = v'(t) = p''(t): a(t)=6t−6a(t) = 6t - 6. To find when the acceleration is zero, set a(t)=0a(t) = 0: 6t−6=06t - 6 = 0. Solving for tt gives 6t=66t = 6, so t=1t=1. This is a positive value.

Question 10

The parabola y=ax2+bxy = ax^2 + bx has a tangent line at the point (1,5)(1, 5) with a slope of 88. What is the value of a−ba-b?

  1. −3-3
  2. 11 (correct answer)
  3. 22
  4. 33
Explanation: First, since the point (1,5)(1, 5) is on the curve, its coordinates must satisfy the equation: 5=a(1)2+b(1)5 = a(1)^2 + b(1), which simplifies to a+b=5a+b=5. Second, the slope of the tangent line is given by the derivative, y′=2ax+by' = 2ax + b. We are given that the slope at x=1x=1 is 8, so y′(1)=2a(1)+b=8y'(1) = 2a(1) + b = 8, which simplifies to 2a+b=82a+b=8. We now have a system of two linear equations: (1) a+b=5a+b=5 and (2) 2a+b=82a+b=8. Subtracting equation (1) from equation (2) gives (2a+b)−(a+b)=8−5(2a+b) - (a+b) = 8-5, which results in a=3a=3. Substituting a=3a=3 into equation (1) gives 3+b=53+b=5, so b=2b=2. The question asks for the value of a−ba-b, which is 3−2=13-2=1.

Question 11

For the function f(x)=x3−6xf(x) = x^3 - 6x, at what positive value of cc is the instantaneous rate of change equal to the average rate of change over the interval [0,3][0, 3]?

  1. 33
  2. 2\sqrt{2}
  3. 3\sqrt{3} (correct answer)
  4. 22
Explanation: First, calculate the average rate of change over the interval [0,3][0, 3]: f(3)−f(0)3−0=(33−6(3))−(03−6(0))3=27−183=93=3\frac{f(3)-f(0)}{3-0} = \frac{(3^3 - 6(3)) - (0^3 - 6(0))}{3} = \frac{27-18}{3} = \frac{9}{3} = 3. Next, find the instantaneous rate of change by taking the derivative: f′(x)=3x2−6f'(x) = 3x^2 - 6. We need to find the value cc where the instantaneous rate of change equals the average rate of change, so we set f′(c)=3f'(c) = 3. This gives the equation 3c2−6=33c^2 - 6 = 3. Solving for cc: 3c2=93c^2 = 9, so c2=3c^2 = 3, which means c=±3c = \pm\sqrt{3}. Since the question asks for the positive value, the answer is c=3c=\sqrt{3}.

Question 12

What is the equation of the line normal to the graph of f(x)=x2xf(x) = x^2\sqrt{x} at x=4x=4?

  1. y=20x−48y = 20x - 48
  2. y=−20x+112y = -20x + 112
  3. y=−120x+32y = -\frac{1}{20}x + 32
  4. y=−120x+1615y = -\frac{1}{20}x + \frac{161}{5} (correct answer)
Explanation: First, simplify the function: f(x)=x2⋅x1/2=x5/2f(x) = x^2 \cdot x^{1/2} = x^{5/2}. Next, find the derivative to get the slope of the tangent line: f′(x)=52x3/2f'(x) = \frac{5}{2}x^{3/2}. Evaluate at x=4x=4 to find the tangent slope: mtan=f′(4)=52(4)3/2=52(8)=20m_{tan} = f'(4) = \frac{5}{2}(4)^{3/2} = \frac{5}{2}(8) = 20. The slope of the normal line is the negative reciprocal of the tangent slope, so mnorm=−120m_{norm} = -\frac{1}{20}. Find the y-coordinate of the point on the curve at x=4x=4: f(4)=45/2=(4)5=25=32f(4) = 4^{5/2} = (\sqrt{4})^5 = 2^5 = 32. The point is (4,32)(4, 32). Using the point-slope form, the equation of the normal line is y−32=−120(x−4)y - 32 = -\frac{1}{20}(x - 4). Simplifying to slope-intercept form: y=−120x+420+32=−120x+15+1605=−120x+1615y = -\frac{1}{20}x + \frac{4}{20} + 32 = -\frac{1}{20}x + \frac{1}{5} + \frac{160}{5} = -\frac{1}{20}x + \frac{161}{5}.