I have 3 calc problems that i have not been able to do please help me... i've been on here yesterday as well but no one was able to assist me with these particular ones...
1. If f(x)=5x^3–2/x-4
find f'(x)____
what i got was (4x^2+7x-12)/x^8
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
1. If f(x)=5x^3–2/x-4
find f'(x)____
=============================
I assume you mean
(5 x^3 -2)
----------
(x-4)
(x-4)(15 x^2) - (5 x^3-2)(1)
----------------------------
x^2 - 8 x + 16
15 x^3 -60 x^2 - 5 x^3 +2
-------------------------
x^2 - 8 x + 16
10 x^3 -60 x^2 +2
-----------------
x^2 - 8 x + 16
what i got was (4x^2+7x-12)/x^8
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
===========================
s' = 3 t^2 - 8 t - 7 = 7
3 t^2 - 8 t -14 = 0
t = [ 8 +/- sqrt (64 +84) ] /6
= [8 +/- 12.17]/6
=3.36
==============================
3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
sqrt (x) (4x+7) - (2x^2 + 7 x +3)(1/2x)sqrt x
----------------------------------------
x
2 x sqrt (x)(4x+7) - (2 x^2+7x+3) sqrt x
---------------------------------------
x
(sqrt x)[8 x^2 +14 x - 2 x^2 -7 x - 3]
--------------------------------------
x
(sqrt x/x) (6 x^2 + 7 x -3)
6 x^(3/2) + 7 x^(1/2) - 3 x^-(1/2)
======================================
i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
I don't exactly know why but all of the answers were wrong.
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
===========================
s' = 3 t^2 - 8 t - 7 = 10
3 t^2 - 8 t -17 = 0
t = [ 8 +/- sqrt (64 + 204) ] /6
= [8 +/- 16.4]/6
=4.94
Sure! I'll be glad to help you with these calculus problems. Let's go through each one step by step:
1. To find the derivative of the function f(x) = 5x^3 - 2/(x - 4), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), then its derivative is given by (g'(x)h(x) - g(x)h'(x))/[h(x)]^2.
Applying the quotient rule to our function, we have:
f'(x) = [(3)(5x^2)(x - 4) - (5x^3 - 2)(1)] / [(x - 4)^2]
= (15x^2(x - 4) - (5x^3 - 2)) / (x - 4)^2
= (15x^3 - 60x^2 - 5x^3 + 2) / (x - 4)^2
= (-10x^3 - 60x^2 + 2) / (x - 4)^2
So, the derivative of f(x) is (-10x^3 - 60x^2 + 2) / (x - 4)^2.
2. To find when the particle reaches a velocity of 10 m/s, we need to find the time(s) when the particle's velocity is equal to 10 m/s. Velocity is the derivative of position with respect to time. In other words, v(t) = s'(t), where v(t) is the velocity function and s(t) is the position function.
Given the position function s(t) = t^3 - 4t^2 - 7t, we can find the velocity function by taking its derivative:
v(t) = s'(t) = (3t^2 - 8t - 7)
Now, we can set v(t) equal to 10 and solve for t:
10 = 3t^2 - 8t - 7
0 = 3t^2 - 8t - 17
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-(-8) ± √((-8)^2 - 4(3)(-17))) / (2(3))
= (8 ± √(64 + 204)) / 6
= (8 ± √268) / 6.
Since we are only interested in the time when t >= 0, we can exclude the negative solution:
t = (8 + √268) / 6.
Therefore, the particle reaches a velocity of 10 m/s at t = (8 + √268) / 6 seconds.
3. To find the derivative of the function f(x) = (2x^2 + 7x + 3) / √x, we can use the quotient rule again.
Applying the quotient rule:
f'(x) = [(2)(2x + 7)(√x) - (2x^2 + 7x + 3)(1/2)(x^(-1/2))] / (√x)^2
= (4x + 14)(√x) - (1/2)(2x^2 + 7x + 3)(x^(-1/2)) / x.
Simplifying further, we get:
f'(x) = (4x + 14)(√x) - (x^(-1/2))(x^2 + 7x + 3) / 2x
= (4x + 14)(√x) - (x^(1/2))(x^2 + 7x + 3) / 2x
= (4x + 14)(√x) - (x^3/2 + 7x^(3/2) + 3x^(1/2)) / 2x.
So, the derivative of f(x) is (4x + 14)(√x) - (x^3/2 + 7x^(3/2) + 3x^(1/2)) / 2x.
I hope this helps! Let me know if you have any further questions.