CALCULUS HELP PLS!?

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Calculus help needed!?
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I am taking my first calculus course and I honestly have no idea what I am doing. I took a few approaches to the problems but I'm doing something wrong. Explanations would be highly appreciated so I could see how you did the problems.

Question 1:
Consider the function : f(x)=
(3x^3 + |x|)/x
(a) Write f(x) as a piecewise function.
(b) Sketch the graph of f(x).
(c) Is the function f(x) defined at x = 0? Justify.
(d) Find the slope of the tangent line to the curve at the point x = 3

Question 2:
If an arrow is shot upward on the moon with a velocity of 58 m/s, its height in meters after t
seconds is given by
h(t) = 58 t − 0.83 t

(a) Give an expression for the average velocity of the arrow on the time interval [1, 1 + a],
and simplify the resulting expression.
(b) Find the average velocity on the given time intervals:
(i) [1, 1.1]
(i) [1, 1.01]
(i) [1, 1.001]
(c) Use the definition of the derivative to find the instantaneous velocity after one second

Thank you to anyone who helps.

Answer 1

Question 1:

y = |x| is a v shaped function like this.

http://www.wolframalpha.com/input/?i=f(x)%3D+%7Cx%7C,+(x+from+-2+to+2)



y = |x|/x become a step function like this.

http://www.wolframalpha.com/input/?i=f(x)%3D+%7Cx%7C%2Fx,+(x+from+-2+to+2)

with values -1 for x < 0 and + 1 for x > 0



f(x) = (3x^3 + |x|)/x = 3x^2 + |x|/x

(b) The sketch is two halves of a parabola, each half down or up like this.

http://www.wolframalpha.com/input/?i=f(x)%3D+3x%5E2+%2B+%7Cx%7C%2Fx,+(x+from+-2+to+2)



(a) As a piecewise function it is:-

f(x) = 3x^2 - 1 for x < 0

f(x) = 3x^2 + 1 for x > 0



(c) f(0)= [3(0)^3 + |0|]/x0

f(x) is indeterminate at x = 0.



(d) f (x) = 6x, so, when x = 3 the slope is 18



Question 2: Gravitational field strength on the Moon is about 1.6 m/s^2, so it seems likely that your equation should be h(t) = 58t - 0.83t^2



Note: Velocity calculations easier if allowed to use dh/dt, but the structure of the question sugggests we should not consider that until later.



(a)

h(1) = 58 - 0.83 = 57.17

h(1 + a) = 58 + 58a - 0.83(1 + 2a + a^2) = 57.17 + 56.34a – 0.83a^2

Increase in height = 56.34a – 0.83a^2

Increase in time = a

Average velocity V = 56.34 - 0.83a



(b) These are the average velocities for the given time intervals:

(i) a = 0.1, so V = 56.34 - 0.083 = 56.257 m/s

(ii) a = 0.01, so V = 56.34 - 0.0083 = 56.3317 m/s

(iii) a = 0.001, V = 56.34 -0.00083 = 56.33917 m/s

We are supposed to observe that as a gets smaller, V approaches 56.34



(c) We are directed to find an expression for instantaneous velocity after one second by using the definition of the derivative which (adapted) is

v(t) = dh(t)/dt = limit of s -> 0 of [h(t + s) - h(t)]/s ……………..(1)

h(t + s) = 58t + 58s - 0.83(t^2 + 2ts + s^2)

h(t) = 58t - 0.83t^2

[h(t + s) - h(t)]/s = [58s – 0.83(2ts + s^2]/s = 58 – 1.66t - 0.83 s

limit of s -> 0 of that is 58 – 1.66t



Compare dh/dt = 58 – 1.66t, (using power rule), evaluated at t = 1



In either case, when t = 1, instantaneous velocity is 56.34 m/s

Answer 2

1a) The first problem is an algebra problem. By the time you get to a calculus class, you are supposed to know what the absolute value function does. It has one behavior for x < 0 and a different behavior for x > 0. Thus, x=0 is where the break lies between one piece of the function definition and the other.

.. f(x) = {x < 0: (3x^3 -x)/x

.. .. .. .. {x > 0: (3x^3 +x)/x

Obviously, this simplifies to

.. f(x) = {x < 0: 3x^2 -1

.. .. .. .. {x > 0: 3x^2 +1

.. .. .. .. {x = 0: undefined . . . . . . the value of |x|/x is not defined at x=0



1b) See the source link for a graph



1c) The value of |x|/x is -1 for x<0, +1 for x>0. The limit value of this ratio at x=0 is different from the right than from the left, so is said not to exist. If the value of |x|/x does not exist at x=0, then f(x) does not exist at x=0.



1d) x=3 is greater than zero, so the function 3x^2 +1 applies. The slope of this is 6x, which has the value of 6*3 = 18 at x=3. The slope of the tangent at x=3 is 18.





2) We suspect your equation for height should be

.. h(t) = 58t -0.83t^2 . . . . . . since Moon's gravity is about 1/6 that of Earth's

2a) The average velocity is defined as

.. Vavg = (h(1+a) -h(1))/((1 +a) -1)

.. = ((58(1 +a) -0.83(1 +a)^2) - (58*1 -0.83*1^2))/a

.. = (58a -0.83*(2a +a^2))/a

.. = 58 -0.83(2 +a)

.. = 56.34 -0.83a



2bi) a=0.1, Vavg = 56.34 -0.83*0.1 = 56.257 . . . m/s

2bii) a=0.01, Vavg = 56.34 -0.83*.01 = 56.3317 . . . m/s

2biii) a=0.001, Vavg = 56.34 -0.83*.001 = 56.33917 . . . m/s



2c) dh/dt = lim[a→0] Vavg = 56.34 -0.83*0 = 56.34 . . . m/s (at t=1)



_____

There's no point to trying wrong ways to work problems like these. The purpose of your text is to teach you the right way to work them. If you don't understand what it is telling you, ask questions. (It will assume you understand the rules and terminology of algebra. If you don't, that is a different problem.)



For position, velocity, and acceleration problems, you are expected to know that the velocity is the rate of change of position, and the acceleration is the rate of change of velocity. The average velocity is

.. Vavg = (change in position)/(time in which that change occurred)

Since h(t) gives position, the change in position is

.. (change in position) = (position at some time) - (position at some earlier time)

In this problem, you are given the times as "1" and "1+a", so you have

.. (change in position) = h(1+a) -h(1)

.. (time in which that change occurred) = (1+a) -(1)

Answer 3

　

1.



f(x) = (3x^3 + |x|)/x



(a)



When x < 0, then |x| = −x ----> f(x) = (3x^3 − x) / x = 3x² − 1

When x > 0, then |x| = x ----> f(x) = (3x^3 + x) / x = 3x² + 1



f(x) = 3x² − 1 ..... x < 0

f(x) = 3x² + 1 ..... x > 0



(b)



https://www.desmos.com/calculator/19v35fp0in



(c)



No, because when x = 0, we get division by 0



(d) Find the slope of the tangent line to the curve at the point x = 3



x = 3 > 0 ----> f(x) = 3x² + 1 ----> f'(x) = 6x

f(3) = 3(3)² + 1 = 28

f'(3) = 6(3) = 18



Tangent line passes through point (3,28) and has slope = 18



y − 28 = 18 (x − 3)

y = 18x − 26



——————————————————————————————



2.



Function doesn't look right. It's most probably: h(t) = 58 t − 0.83 t^2



(a)



Average velocity

= (h(1+a) − h(1)) / ((1+a) − 1)

= ((58 (1+a) − 0.83 (1+a)²) − (58 − 0.83)) / a

= (58 + 58a − 0.83 − 1.66a − 0.83a² − 58 + 0.83) / a

= (56.34a − 0.83a²) / a

= 56.34 − 0.83a



(b)



(i) [1, 1.1] ----> a = 0.1 ----> 56.34 − 0.83(0.1) = 56.3317

(i) [1, 1.01] ----> a = 0.01 ----> 56.34 − 0.83(0.01) = 56.3317

(i) [1, 1.001] ----> a = 0.001 ----> 56.34 − 0.83(0.001) = 56.33917



(c)



lim[a→0] (h(1+a) − h(1)) / ((1+a) − 1)

= lim[a→0] (56.34 − 0.83a)

= 56.34