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If V = 4/3 πr³, at what rate in cubic units is V increasing when r = 10 and dr/dt = 0.01?
To find the rate at which the volume V is increasing, we need to differentiate the volume formula V = (4/3) π r³ with respect to time t. dV/dt = 4 π r² (dr/dt) Now, plug in the values: r = 10 and (dr/dt) = 0.01: dV/dt = 4 π (10)² (0.01) dV/dt = 4 π × 100 × 0.01 = 4 π Thus, the rate at which the voluRead more
To find the rate at which the volume V is increasing, we need to differentiate the volume formula V = (4/3) π r³ with respect to time t.
dV/dt = 4 π r² (dr/dt)
Now, plug in the values: r = 10 and (dr/dt) = 0.01:
dV/dt = 4 π (10)² (0.01)
dV/dt = 4 π × 100 × 0.01 = 4 π
Thus, the rate at which the volume is increasing is 4π cubic units.
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The function f(x) = tan x – x
The function f(x) = tan x - x increases or decreases throughout, and the only way to find this is by analyzing the derivative of f(x). Now, differentiate f(x) with respect to x: f'(x) = d/dx(tan x) - d/dx(x) = sec² x - 1 So, f'(x) = sec² x - 1 = tan² x Since tan² x is always non-negative for all reaRead more
The function f(x) = tan x – x increases or decreases throughout, and the only way to find this is by analyzing the derivative of f(x).
Now, differentiate f(x) with respect to x:
f'(x) = d/dx(tan x) – d/dx(x) = sec² x – 1
So,
f'(x) = sec² x – 1 = tan² x
Since tan² x is always non-negative for all real values of x, f'(x) ≥ 0 for all x where tan x is defined.
Therefore, f(x) is always increasing where it is defined, but has vertical asymptotes at x = (π/2) + nπ, where n is any integer.
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For all real values of x, the function f(x) = |x| is
The function f(x) = |x| is neither increasing nor decreasing throughout its domain since: - For x ≥ 0, the function is increasing (since f(x) = x for nonnegative x). - For x < 0, the function is decreasing (since f(x) = -x for negative x). The function is neither strictly increasing nor strictlyRead more
The function f(x) = |x| is neither increasing nor decreasing throughout its domain since:
– For x ≥ 0, the function is increasing (since f(x) = x for nonnegative x).
– For x < 0, the function is decreasing (since f(x) = -x for negative x).
The function is neither strictly increasing nor strictly decreasing at x = 0 because of a sharp "corner."
Hence f(x) = |x| is neither increasing nor decreasing on its entire domain.
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The function f(x) = x³ – 3x² – 9x – 7 has a stationary point at
To find the stationary points of the function f(x) = x³ - 3x² - 9x - 7, one would first have to compute the derivative of the function and set the function equal to zero. Stationary points occur at places where the derivative is equal to zero. First, differentiate f(x): f'(x) = d/dx(x³ - 3x² - 9x -Read more
To find the stationary points of the function f(x) = x³ – 3x² – 9x – 7, one would first have to compute the derivative of the function and set the function equal to zero. Stationary points occur at places where the derivative is equal to zero.
First, differentiate f(x):
f'(x) = d/dx(x³ – 3x² – 9x – 7) = 3x² – 6x – 9
Now, put f'(x) = 0 to get the stationary points:
3x² – 6x – 9 = 0
Divide the equation by 3:
x² – 2x – 3 = 0
Factor the quadratic equation:
(x – 3)(x + 1) = 0
Therefore, the solutions are x = 3 and x = -1.
Hence, the stationary points are at x = -1 and x = 3.
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The value of b for which the function f(x) = x + cos x + b is strictly decreasing over R is
In order to determine the value of b for which the function f(x) = x + cos x + b is strictly decreasing over ℝ, we start by analyzing the derivative of the function. Let us differentiate f(x): f'(x) = d/dx(x + cos x + b) = 1 - sin x Hence for being strictly decreasing, the derivative must be negativRead more
In order to determine the value of b for which the function f(x) = x + cos x + b is strictly decreasing over ℝ, we start by analyzing the derivative of the function. Let us differentiate f(x):
f'(x) = d/dx(x + cos x + b) = 1 – sin x
Hence for being strictly decreasing, the derivative must be negative for all values of x; i.e., f'(x) < 0.
1 – sin x 1
But the function sin x can never be more than 1 for any real number x. Thus, b cannot be taken such that for all real number x, f'<0.
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