Power Rule - Calculus 1

What is the equation of the line normal to the graph of $f(x) = x^2\sqrt{x}$ at $x=4$ ?
Answer: $y = -\frac{1}{20}x + \frac{161}{5}$ First, simplify the function: $f(x) = x^2 \cdot x^{1/2} = x^{5/2}$ . Next, find the derivative to get the slope of the tangent line: $f'(x) = \frac{5}{2}x^{3/2}$ . Evaluate at $x=4$ to find the tangent slope: $m_{tan} = f'(4) = \frac{5}{2}(4)^{3/2} = \frac{5}{2}(8) = 20$ . The slope of the normal line is the negative reciprocal of the tangent slope, so $m_{norm} = -\frac{1}{20}$ . Find the y-coordinate of the point on the curve at $x=4$ : $f(4) = 4^{5/2} = (\sqrt{4})^5 = 2^5 = 32$ . The point is $(4, 32)$ . Using the point-slope form, the equation of the normal line is $y - 32 = -\frac{1}{20}(x - 4)$ . Simplifying to slope-intercept form: $y = -\frac{1}{20}x + \frac{4}{20} + 32 = -\frac{1}{20}x + \frac{1}{5} + \frac{160}{5} = -\frac{1}{20}x + \frac{161}{5}$ .
The curve given by the equation $y = \frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x + 1$ has horizontal tangent lines at two distinct points. What is the sum of the x-coordinates of these points?
Answer: $3$ A horizontal tangent line occurs where the derivative is equal to zero. First, find the derivative of the function using the Power Rule: $y' = \frac{d}{dx}(\frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x + 1) = x^2 - 3x - 4$ . Set the derivative equal to zero to find the x-coordinates: $x^2 - 3x - 4 = 0$ . This quadratic equation can be factored as $(x-4)(x+1) = 0$ , which gives solutions $x=4$ and $x=-1$ . The sum of these x-coordinates is $4 + (-1) = 3$ .
Let $f(x) = \frac{x^3 - \sqrt{x}}{x}$ . What is the value of $f'(4)$ ?
Answer: $\frac{129}{16}$ First, rewrite the function $f(x)$ by splitting the fraction: $f(x) = \frac{x^3}{x} - \frac{\sqrt{x}}{x} = x^2 - x^{1/2}x^{-1} = x^2 - x^{-1/2}$ . Now, apply the Power Rule to find the derivative: $f'(x) = 2x - (-\frac{1}{2})x^{-3/2} = 2x + \frac{1}{2}x^{-3/2}$ . Finally, evaluate the derivative at $x=4$ : $f'(4) = 2(4) + \frac{1}{2}(4)^{-3/2} = 8 + \frac{1}{2} \cdot \frac{1}{(4^{1/2})^3} = 8 + \frac{1}{2} \cdot \frac{1}{2^3} = 8 + \frac{1}{2} \cdot \frac{1}{8} = 8 + \frac{1}{16} = \frac{128}{16} + \frac{1}{16} = \frac{129}{16}$ .
The parabola $y = ax^2 + bx$ has a tangent line at the point $(1, 5)$ with a slope of $8$ . What is the value of $a-b$ ?
Answer: $1$ First, since the point $(1, 5)$ is on the curve, its coordinates must satisfy the equation: $5 = a(1)^2 + b(1)$ , which simplifies to $a+b=5$ . Second, the slope of the tangent line is given by the derivative, $y' = 2ax + b$ . We are given that the slope at $x=1$ is 8, so $y'(1) = 2a(1) + b = 8$ , which simplifies to $2a+b=8$ . We now have a system of two linear equations: (1) $a+b=5$ and (2) $2a+b=8$ . Subtracting equation (1) from equation (2) gives $(2a+b) - (a+b) = 8-5$ , which results in $a=3$ . Substituting $a=3$ into equation (1) gives $3+b=5$ , so $b=2$ . The question asks for the value of $a-b$ , which is $3-2=1$ .
For the function $f(x) = x^3 - 6x$ , at what positive value of $c$ is the instantaneous rate of change equal to the average rate of change over the interval $[0, 3]$ ?
Answer: $\sqrt{3}$ First, calculate the average rate of change over the interval $[0, 3]$ : $\frac{f(3)-f(0)}{3-0} = \frac{(3^3 - 6(3)) - (0^3 - 6(0))}{3} = \frac{27-18}{3} = \frac{9}{3} = 3$ . Next, find the instantaneous rate of change by taking the derivative: $f'(x) = 3x^2 - 6$ . We need to find the value $c$ where the instantaneous rate of change equals the average rate of change, so we set $f'(c) = 3$ . This gives the equation $3c^2 - 6 = 3$ . Solving for $c$ : $3c^2 = 9$ , so $c^2 = 3$ , which means $c = \pm\sqrt{3}$ . Since the question asks for the positive value, the answer is $c=\sqrt{3}$ .
Let $g(x) = \frac{2}{x^3} + 3x^2$ . Find the value of the second derivative, $g''(1)$ .
Answer: $30$ First, rewrite the function in terms of powers: $g(x) = 2x^{-3} + 3x^2$ . Now, find the first derivative using the Power Rule: $g'(x) = 2(-3)x^{-4} + 3(2)x^1 = -6x^{-4} + 6x$ . Next, find the second derivative by differentiating $g'(x)$ : $g''(x) = -6(-4)x^{-5} + 6 = 24x^{-5} + 6$ . Finally, evaluate $g''(x)$ at $x=1$ : $g''(1) = 24(1)^{-5} + 6 = 24 + 6 = 30$ .
The line tangent to the curve $y = x^2 + c$ at $x=2$ passes through the point $(1, 1)$ . What is the value of $c$ ?
Answer: $1$ 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 $x = 2$ , you need both a point on the curve and the slope at that point. The derivative $y' = 2x$ gives the slope, so at $x = 2$ , the slope is $2(2) = 4$ . The point on the curve is $(2, 4 + c)$ since $y = 2^2 + c = 4 + c$ . Using point-slope form with slope $m = 4$ and point $(2, 4 + c)$ : $y - (4 + c) = 4(x - 2)$ $y = 4x - 8 + 4 + c = 4x - 4 + c$ Since this tangent line passes through $(1, 1)$ , substitute these coordinates: $1 = 4(1) - 4 + c$ $1 = 4 - 4 + c$ $1 = c$ Therefore, $c = 1$ , which is choice D. Choice A ( $c = 0$ ) would give tangent line $y = 4x - 4$ , which equals 0 at $x = 1$ , not 1. Choice B ( $c = -3$ ) gives $y = 4x - 7$ , which equals $-3$ at $x = 1$ . Choice C ( $c = -1$ ) gives $y = 4x - 5$ , which equals $-1$ at $x = 1$ . 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.
Let $f(x) = \frac{x^a}{x^b}$ where $a = 3/2$ and $b = -1/2$ . If $g(x)$ is the derivative of $f(x)$ , what is $g(16)$ ?
Answer: $32$ First, simplify the function $f(x)$ using the rules of exponents: $f(x) = x^{a-b}$ . Substitute the given values of $a$ and $b$ : $a-b = \frac{3}{2} - (-\frac{1}{2}) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$ . So, the function simplifies to $f(x) = x^2$ . The function $g(x)$ is the derivative of $f(x)$ . Using the Power Rule, $g(x) = f'(x) = 2x$ . Finally, evaluate $g(x)$ at $x=16$ : $g(16) = 2(16) = 32$ .
The position of a particle moving along the x-axis is given by $p(t) = t^3 - 3t^2 - 9t + 1$ . For what positive value of $t$ is the particle's acceleration equal to zero?
Answer: $1$ 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 $v(t) = p'(t)$ : $v(t) = 3t^2 - 6t - 9$ . Next, find the acceleration function $a(t) = v'(t) = p''(t)$ : $a(t) = 6t - 6$ . To find when the acceleration is zero, set $a(t) = 0$ : $6t - 6 = 0$ . Solving for $t$ gives $6t = 6$ , so $t=1$ . This is a positive value.
Let $h(x) = (x^2+1)(\frac{1}{x} + x)$ . Find the slope of the tangent line to the graph of $h(x)$ at $x=-1$ .
Answer: $4$ To find the slope of the tangent line, we need to find the derivative $h'(x)$ . It is easiest to first expand and simplify the expression for $h(x)$ : $h(x) = x^2(\frac{1}{x}) + x^2(x) + 1(\frac{1}{x}) + 1(x) = x + x^3 + x^{-1} + x = x^3 + 2x + x^{-1}$ . Now, apply the Power Rule to find the derivative: $h'(x) = 3x^2 + 2 - 1x^{-2} = 3x^2 + 2 - \frac{1}{x^2}$ . Finally, evaluate the derivative at $x=-1$ to find the slope: $h'(-1) = 3(-1)^2 + 2 - \frac{1}{(-1)^2} = 3(1) + 2 - \frac{1}{1} = 3 + 2 - 1 = 4$ .
Let $f(x) = ax^n + bx^{n-1}$ where $a$ and $b$ are non-zero real numbers and $n$ is an integer greater than 2. What is the smallest integer $k$ for which the $k$ -th derivative, $f^{(k)}(x)$ , is a non-zero constant?
Answer: $n$ Consider the term $ax^n$ . Each time we differentiate, the power of $x$ decreases by 1. After differentiating $n$ times, the term becomes $a \cdot n! \cdot x^0 = a \cdot n!$ , which is a non-zero constant. Now consider the term $bx^{n-1}$ . After differentiating $n-1$ times, it becomes $b \cdot (n-1)!$ , which is a non-zero constant. After differentiating $n$ times, this term becomes 0. Therefore, the $n$ -th derivative of the entire function is $f^{(n)}(x) = a \cdot n! + 0 = a \cdot n!$ . This is a non-zero constant. For any $k < n$ , at least the first term will still contain a power of $x$ , so $f^{(k)}(x)$ will not be a constant. For $k=n+1$ , the derivative will be 0. Thus, the smallest integer $k$ for which the derivative is a non-zero constant is $k=n$ .
Let $f(x) = x^k$ and $g(x) = x^{k+2}$ , where $k$ is a positive integer. If $f'(2) = g'(1)$ , what is the value of $k$ ?
Answer: $2$ 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, $f'(x) = kx^{k-1}$ and $g'(x) = (k+2)x^{k+1}$ . Now we can evaluate each derivative at the specified points: $f'(2) = k \cdot 2^{k-1}$ and $g'(1) = (k+2) \cdot 1^{k+1} = k+2$ . Since $f'(2) = g'(1)$ , we have the equation $k \cdot 2^{k-1} = k+2$ . Let's test the answer choices systematically. For choice A ( $k=1$ ): $1 \cdot 2^0 = 1 \cdot 1 = 1$ , but $k+2 = 3$ . Since $1 \neq 3$ , this is incorrect. For choice B ( $k=4$ ): $4 \cdot 2^3 = 4 \cdot 8 = 32$ , but $k+2 = 6$ . Since $32 \neq 6$ , this is incorrect. For choice C ( $k=3$ ): $3 \cdot 2^2 = 3 \cdot 4 = 12$ , but $k+2 = 5$ . Since $12 \neq 5$ , this is incorrect. For choice D ( $k=2$ ): $2 \cdot 2^1 = 2 \cdot 2 = 4$ , and $k+2 = 4$ . Since $4 = 4$ , 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.

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