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# This question uses the function f(x) = -2x^3 - 9x^2 + 24x +

### Customer Question

This question uses the function f(x) = -2x^3 - 9x^2 + 24x + 40
Find the stationary points of this function

B. Using the strategy to apply the first derivative test, classify the left hand stationary point found in the first question

Then using second derivative test classify the right hand stationary point found in first question

C. Find the y coordinate of each of the stationary points on the graph of the function f(x), and evaluate f(0)

D. Draw a graph of the function f(x)

2a. Simplify the answers where appropriate
Write down the derivative of each of the functions:
F(x) = x^4 and g(x) = sin(5x)

Hence by using product rule, differentiate the function
K(x) = x^4sin(5x)

Write down the derivative of each of the functions f(t) = e^6t and g(t) = t^3 + 8

Hence by using quotient rule, differentiate the function
K(t) = e^6t/t^3 + 8 (t > -2)

Write down derivative of the function f(x) = 6cos(1/3x)

Hence using the composite rule differentiate the function k(x) = ln (6cos(1/3x)) (-3/2 delta < x 0)
h(u) = cos^2 (6u)

B. Evaluate indefinite integral with 8 at top and 6 at bottom x(4 + 7x^2)dx
C. Write down a definite integral that will give the value of the area under the curve y=x^2sin(3/4x) between x = 1/3 delta and x = delta
( do not evaluate the integral by hand)

Use mathcad to find the area described in previous question giving answer to 4 decimal places and give print out of the working

4. A rocket is modelled by a particle that moves along a vertical line. From launch the rocket rises until it's motor cuts out after 17 seconds. At this time it has reached a height of 580 metres above launch pad and attained an upward velocity of 120ms-1. from this time on the rocket has a constant upward acceleration of -10ms-2 ( due to gravity effect)

Choose the s axis ( for position of particle that represents the rocket) to point upwards with origin at launch pad. Take t = 0 to be the time when rocket motor cuts
What is the max height above launch pad reached by rocket?
How long from launch does rocket take to reach this max height?
After how long from launch does rocket crash onto launch pad?

5. Solve the initial value problem dy/dx = sin(4x)/3+cos(4x). Y = 5 when x = 0

Show where c is an arbitrary constant in this equation
Indefinate integral 2e^2x - 5/(e^2x - 5x)^2/3 dx=3(e^2x - 5x)^1/3 + c

Hence find it's implicit form, the general solution of the differential equation
Dy/dx = 2y^1/2(2e^2x -5)/3(e^2x -5x)^2/3 (y>0)

Find corresponding particular solution (in implicit form) that satisfies the initial condition y=1 when x = 0

Find the explicit form of this particular solution

What is the value of y given by this particular solution when x =2? Give your answer to 4 sig figs
Submitted: 5 years ago.
Category: Calculus and Above
Expert:  Angela--Mod replied 5 years ago.
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Customer: replied 5 years ago.
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Expert:  Calc101 replied 5 years ago.

So, a stationary point is the same as a critical point (if you know that term better). Basically we need to find where the derivative is equal to 0. So, let's differentiate f(x).

f'(x)=-6x^2-18x+24

Now, set the derivative equal to 0, and you can solve for what the stationary points are.

0=-6x^2-18x+24
0=x^2+3x-4
0=(x+4)(x-1)

Therefore, x=-4 and x=1 are your stationary points.

B. To use the first derivative test, we will test what the derivative does to the left of x=-4 and to the right of x=-4.

To the left of x=-4, f'(x)0.

When the slope goes from negative to positive, we have a relative minimum.

For the second derivative test, we must of course first find the the second derivative.

f''(x)=12x-18

Now, we test f''(1). If it is less than 0, we have a maximum, and if it is greater than 0, we have a minimum.

f''(1)=-6

Therefore, we have a relative maximum.

We can also verify this using the first derivative test, since to the left of x=1 (and to the right of x=-4), f'(x)>0 and to the right of x=1, f'(x)<0, which means we have a relative maximum.

C. f(0)=53
f(-4)=-72
f(1)=40

2.

A. f'(x)=4x^3
g'(x)=5cos(5x) (this involves the chain rule where we must take the derivative of the outer function (sin) and then multiply it by the derivative of the inside function (5x) which is 5).

The derivative using the product rule is fg'+gf'.

So, we take x^4(5cos(5x))+sin(5x)(4x^3)

Now for the second one:

f'(t)=6e^6t, since we take the derivative of e^t which is just e^t and multiply it by the derivative of 6t (which is 6). This is basically u substitution where u=6t and du=6.

g'(t)= 3t^2

The quotient rule is (gf'-fg')/g^2 PLEASE NOTE: In the quotient rule the top (gf'-fg') matters in terms of order. The product rule does not. It can be written as fg'+gf' or gf'+fg'.

So, we take ((t^3+8)(6e^6t)-(e^6t)(3t^2))/(t^3+8)^2

Now to the next part,

f(x)=6cos(1/3x)

This utilizes the chain rule again.

We take the derivative of the outside, 6cos, which is -6sin (please don't forget that cosine has the negative when you take the derivative, whereas the derivative of sine does not). Then take the derivative of the inside ((1/3)x) which is 1/3.

We get, -6sin((1/3)x)*1/3, which is -2sin((1/3)x).

So, using the composite rule (AKA chain rule), we basically take the derivative of the outside, ln(u) (where u is the inside function), which is 1/u du.

du is the derivative of the inside, namely, -2sin((1/3)x).

B. (First of all, the instructions must be wrong here, since an indefinite integral has no bounds. Definite integration involves the bounds from 6 to 8).

So, this involves u substitution. Make u the function that will result in giving you what is leftover with the dx.

So, let's try u=4+7x^2 since that will leave us with something times xdx. du is the derivative of u. So, du=14x, which means du/14=xdx.

So, we now have (1/14)int(u du,6,8). We put the constant that goes with du in front of the integral when we integrate. The integral of u, is u^2/2, which when we replace is (4+7x^2)/2 from 6 to 8.

Start with the upper bound (8), that gives us, 226. Now plug in the lower bound (6), which gives us 128. Now subtract the lower bound's value from the upper bound's value. 226-128=98.