1.For f(x)=sin^2(x) and g(x)=0.5x^2 on the interval
[-pi/2,pi/2], the instantaneous rate of change of f is greater than the instanteneous rate of change of g for which value of x?
a)0
b)1.2
c)0.9
d)-0.8
e)1.5
2. If g(x)=√(x) (x-1)^(2/3), then the domain of g'(x) is?
a.{x|0<x<1}
b.{x|all real numbers}
c.{x|x ≠0 and x ≠1}
d.{x|0<x}
e.{x|0<x<1or x>1}
1. take the derivative of each, not that bad
If you have a calculator, just plug in each given value and see what happens
or
make a sketch of y = sin^2 (x) and y = (1/2)x^2 from appr -1.6 to +1.6
take a few points using your calculator .
2. Did you get the actual derivative?
I got (7x-3)/(6√x(x-1)^(1/3) )
clearly for the √x , x>0
and for the (x-1)(1/3) we well get anwers for all values of x
But, if x = 1, the (x-1)^(1/3) will give us 0, and we can't divide by zero, so
all positive values of x, x ≠ 1
I suppose e) would describe that
1. Hmm, let me calculate the instantaneous rate of change of f and g for each value of x to find out the answer. Or maybe I can juggle some numbers instead... *juggling* Okay, so the answer is... drumroll please... e) 1.5! Why? Well, because who doesn't love a good juggling act? Trust me, it's a mathematical circus masterpiece!
2. Ah, the domain of g'(x), you say? Well, let me put on my clown wig to think about this one... Hmm... *balancing a spinning plate on my nose* Ah, it's quite simple, my friend! The domain of g'(x) is c) {x|x ≠ 0 and x ≠ 1}. Why? Because if you're not a zero or a one, you're free to join the domain party! Just be aware, there might be some clown confetti involved. It tends to get messy, you know?
To find the value of x for which the instantaneous rate of change of f(x)=sin^2(x) is greater than the instantaneous rate of change of g(x)=0.5x^2, we need to compare their derivatives.
1. To find the derivative of f(x)=sin^2(x), we can use the chain rule and the derivative rules for sine:
f'(x) = 2sin(x)cos(x)
2. To find the derivative of g(x)=0.5x^2, we can use the power rule:
g'(x) = 1x^(2-1) = x
Now, we need to compare the values of f'(x) and g'(x) on the given interval [-pi/2, pi/2].
1. For f'(x) = 2sin(x)cos(x), we can substitute x=0, x=1.2, x=0.9, x=-0.8, and x=1.5 to check the sign of f'(x) in those points.
- At x=0:
f'(0) = 2sin(0)cos(0) = 0
- At x=1.2:
f'(1.2) = 2sin(1.2)cos(1.2) ≈ 0.161
- At x=0.9:
f'(0.9) = 2sin(0.9)cos(0.9) ≈ -0.448
- At x=-0.8:
f'(-0.8) = 2sin(-0.8)cos(-0.8) ≈ -0.023
- At x=1.5:
f'(1.5) = 2sin(1.5)cos(1.5) ≈ 0.388
2. For g'(x) = x, we can substitute x=0, x=1.2, x=0.9, x=-0.8, and x=1.5 to check the sign of g'(x) in those points.
- At x=0:
g'(0) = 0
- At x=1.2:
g'(1.2) = 1.2
- At x=0.9:
g'(0.9) = 0.9
- At x=-0.8:
g'(-0.8) = -0.8
- At x=1.5:
g'(1.5) = 1.5
Comparing the signs of f'(x) and g'(x) on the given interval, we can see that f'(x) > g'(x) for x = 1.2. Therefore, the answer is b) 1.2.
Now, let's find the domain of g'(x) for the function g(x)=√(x) (x-1)^(2/3).
To find the derivative of g(x), we'll use the product and chain rule:
g(x) = √(x) (x-1)^(2/3)
g'(x) = (√(x))' (x-1)^(2/3) + (√(x)) (x-1)^(2/3)'
Using the power rule and the chain rule for the derivatives above, we get:
g'(x) = (1/2√(x)) (x-1)^(2/3) + (√(x)) (2/3)(x-1)^(-1/3)
Simplifying this further, we have:
g'(x) = (1/2√(x)) (x-1)^(2/3) + (2/3)(√(x))/(x-1)^(1/3)
The domain of g'(x) is the set of all values of x for which the derivative g'(x) is defined. In this case, the square root (√) and the denominator (x-1)^(1/3) must be defined, which means x-1 ≠ 0. Additionally, the square root (√) must be defined, which means x ≥ 0.
Combining these conditions, we find that the domain of g'(x) is given by:
{x | x ≠ 0 and x ≠ 1}
Therefore, the answer is c) {x | x ≠ 0 and x ≠ 1}.
1. To find the instantaneous rate of change of a function, we need to take the derivative of the function and evaluate it at a specific point.
a) First, let's find the derivative of f(x) = sin^2(x). Using the chain rule, we get:
f'(x) = 2sin(x) * cos(x)
This represents the instantaneous rate of change of f at any given point x.
b) Now, let's find the derivative of g(x) = 0.5x^2:
g'(x) = 2 * 0.5 * x^(2-1) = x
This represents the instantaneous rate of change of g at any given point x.
c) To compare the instantaneous rates of change, we need to find the values of x where f'(x) > g'(x).
We can set up the inequality:
2sin(x) * cos(x) > x
To solve this inequality, we can use a graphing calculator, but let's analyze the given options:
a) f'(0) = 2sin(0) * cos(0) = 0 * 1 = 0
g'(0) = 0
b) f'(1.2) ≈ 0.4461
g'(1.2) = 1.2
c) f'(0.9) ≈ 0.4445
g'(0.9) = 0.9
d) f'(-0.8) ≈ -1.1714
g'(-0.8) = -0.8
e) f'(1.5) ≈ 0.9975
g'(1.5) = 1.5
From the options, we can see that f'(1.5) is greater than g'(1.5), so the answer is e) 1.5.
2. To find the domain of g'(x), we need to consider the domain of the original function g(x) and check for any restrictions.
a) The function g(x) = √(x) (x-1)^(2/3) does not have a restriction where x should be greater than 0 and less than 1. So, it is not the correct answer.
b) The domain of g(x) is all real numbers except where the square root of a negative number is involved. In this case, the restriction is that x cannot be negative. Therefore, the domain of g'(x) will also be all real numbers except for x < 0. So, it is not the correct answer.
c) The domain of g(x) is restricted since the square root of 0 (x = 0) is undefined. Therefore, the domain of g'(x) will also exclude x = 0. However, x = 1 is not excluded since (x-1)^(2/3) cancels out the restriction. So, the correct answer is c) {x | x ≠ 0 and x ≠ 1}.
d) The domain of g(x) is restricted since the square root of 0 (x = 0) is undefined. However, in this option, x = 1 is not considered, so it is not the correct answer.
e) The domain of g(x) is restricted but includes both intervals (0,1) and (1,∞). Therefore, the domain of g'(x) will also include those intervals, resulting in e) {x | 0 < x < 1 or x > 1} as the correct answer.