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1

Answer:

#lim_(xrarr1) 1/(x-1) - 3/(1-x^3)# is #color(green)("undefined")#

Explanation:

Remember the cubic factoring:
#color(white)("XXX")(x^3-1)=(x-1)(x^2+x+1)#

Replace
#color(white)("XXX")1/(x-1)#
with the equivalent
#color(white)("XXX")1/(x-1) * ((x-1)(x^2+x+1))/(x^3-1) = (x^2+x+1)/(x^3-1)#

and replace
#color(white)("XXX")-3/(1-x^3)#
with the equivalent
#color(white)("XXX")-3/(1-x^3)* ( (-1))/( (-1)) = +3/(x^3-1)#

#1/(x-1)-3/(1-x^3)#
becomes
#color(white)("XXX")((x^2+x+1) + 3)/(x^3-1)=(x^2+x+4)/(x^3-1)#

Note that as #xrarr1# the denominator #rarr0# (but the numerator does not, so L'Hopital's rule does not apply)

Since division by zero is undefined, this limit is undefined.

As an aid in verifying this result, here is a graph of #1/(x-1)-3/(1-x^3)#

enter image source here

Notice that whether this limit approaches #-oo# or #+oo# depends upon which side of #1# from which #x# approaches.

1

Answer:

I guessed that you meant #y = sqrt(2x+3)#.

Explanation:

The definition of a derivative is:

#y'(x) = lim_(hto0)(y(x+h)-y(x))/h#

Given:

#y(x) = sqrt(2x+3)#

Then:

#y(x+h) = sqrt(2(x+h)+3)#

Let's find an equation for h.

Square both sides:

#(y(x+h))^2 = 2(x+h)+3#

Use the distributive property:

#(y(x+h))^2 = 2x+ 2h+3#

Please observe that #2x + 3# is the same as #(y(x))^2#:

#(y(x+h))^2 = (y(x))^2 + 2h#

Subtract #(y(x))^2# from both side and flip the equation:

#2h = (y(x+h))^2 - (y(x))^2#

Divide both sides by 2:

#h = ((y(x+h))^2 - (y(x))^2)/2#

Substitute this for h into the definition:

#y'(x) = lim_(hto0)(y(x+h)-y(x))/(((y(x+h))^2 - (y(x))^2)/2)#

Flip the 2 up to the numerartor:

#y'(x) = lim_(hto0)(2(y(x+h)-y(x)))/((y(x+h))^2 - (y(x))^2)#

We know how the difference of two squares factors, #(a^2-b^2)=(a+b)(a-b)#, so we can do this to the denominator:

#y'(x) = lim_(hto0)(2(y(x+h)-y(x)))/((y(x+h) + y(x))(y(x+h) - y(x)))#

Please observe that a common factor cancels:

#y'(x) = lim_(hto0)(2cancel((y(x+h)-y(x))))/((y(x+h) + y(x))cancel((y(x+h) - y(x))))#

Here is the equation with the cancelled factors removed:

#y'(x) = lim_(hto0)2/(y(x+h) + y(x))#

Now it is ok to let #hto0#

#y'(x) = 2/(y(x) + y(x))#

#y'(x) = 2/(2y(x))#

#y'(x) = 1/(y(x))#

#y'(x) = 1/sqrt(2x+3)#

4

Answer:

#implies y = 1/5 cos^(-1) ( pm Ce^((-15/14x^2 tan^-1 (5x) + 3/14x - 3/70 tan^(-1)(5x) )) )#

Explanation:

Is it: #3xtan^-1 (5x) \ dx=7tan(5y) \ dy# ?

If so this means that:

#3xtan^-1 (5x) \ dx - 7tan(5y) \ dy = 0#

We can test if this is exact , ie if there exists (level surface) function #f(x,y) = C# so that:

# df = f_x dx + f_y dy color(red)( = 0 )#

With #f_x = 3xtan^-1 (5x)# and #f_y = - 7tan(5y)#, we test the mixed partials:

#f_(xy) = (partial)/(partial y) (3xtan^-1 (5x)) = 0#

#f_(yx) = (partial)/(partial x) (- 7tan(5y)) = 0#

#f_(xy) = f_(yx)# so the equation is exact and #f(x,y) = C# exists.

Next we can look at:

#f_x = 3xtan^-1 (5x)#

By IBP:

#implies f(x,y) =int 3xtan^-1 (5x) \ dx#

#=int (3/2x^2)^prime tan^-1 (5x) \ dx#

#=3/2x^2 tan^-1 (5x) - int 3/2x^2 (tan^-1 (5x))^prime \ dx#

And as: #(d)/(dx)(tan^(-1)(alpha x)) = alpha/(alpha^2 x^2 + 1)#

#=3/2x^2 tan^-1 (5x) - int 3/2x^2 color(blue)( 5/(25x^2 + 1)) \ dx#

#=3/2x^2 tan^-1 (5x) - 3/10 int (25x^2 + 1 - 1)/(25x^2 + 1) \ dx#

#=3/2x^2 tan^-1 (5x) - 3/10 int 1 - color(blue)( 1/(25x^2 + 1)) \ dx#

And as: #int 1/(alpha^2 x^2 + 1) \ dx= 1/alpha int (d)/(dx)(tan^(-1)(alpha x)) \ dx#

#implies f(x,y) =3/2x^2 tan^-1 (5x) - 3/10x + 3/50 tan^(-1)(5x) + color(red)( eta(y) )#

Because we know that #f_y = - 7tan(5y)#, we can test that against the partial of #f(x,y)#, ie:

# f_y = (partial)/(partial y) ( 3/2x^2 tan^-1 (5x) - 3/10x + 3/50 tan^(-1)(5x) + color(red)( eta(y) ) ) = eta'(y)#

#implies eta' = - 7tan(5y)#

So we integrate again:

# eta(y) = - int 7tan(5y) dy #

We know that: # int tan(alpha y) dy = - 1/alpha ln abs(cos alpha y ) + C#

# implies eta(y) = 7/5 ln abs (cos(5 y)) + C#, where #C# is generic constant term.

This means that:

# f(x,y) =3/2x^2 tan^-1 (5x) - 3/10x + 3/50 tan^(-1)(5x) + 7/5 ln abs (cos(5 y)) = C#

OR:

#7/5 ln abs (cos(5 y)) = - 3/2x^2 tan^-1 (5x) + 3/10x - 3/50 tan^(-1)(5x) +C#

#implies abs cos(5 y) = e^((-15/14x^2 tan^-1 (5x) + 3/14x - 3/70 tan^(-1)(5x) + C ))#

#= C e^((-15/14x^2 tan^-1 (5x) + 3/14x - 3/70 tan^(-1)(5x) ))#

#implies y = 1/5 cos^(-1) ( pm Ce^((-15/14x^2 tan^-1 (5x) + 3/14x - 3/70 tan^(-1)(5x) )) )#

2

Answer:

Please see below. Warning: long answer due to explanatory analysis.

Explanation:

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

Finding the proof

By definition,

#lim_(xrarrcolor(green)(a))color(red)(f(x)) = color(blue)(L)# if and only if

for every #epsilon > 0#, there is a #delta > 0# such that:
for all #x#, #" "# if #0 < abs(x-color(green)(a)) < delta#, then #abs(color(red)(f(x))-color(blue)(L)) < epsilon#.

We have been asked to show that

#lim_(xrarrcolor(green)(-1.5))color(red)((9-4x^2)/(3+2x) = color(blue)(6)#

So we want to make #abs(underbrace(color(red)((9-4x^2)/(3+2x) ))_(color(red)(f(x)) )-underbrace(color(blue)(6))_color(blue)(L))# less than some given #epsilon# and we control (through our control of #delta#) the size of #abs(x-underbrace(color(green)((-1.5))))_color(green)(a))#

We want: #abs((9-4x^2)/(3+2x) )-6) < epsilon#

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

#abs((9-4x^2)/(3+2x) - 6) = abs(((3-2x)(3+2x))/(3+2x) - 6)#

# = abs(((3-2x) - 6)#

#=abs(-2x-3)#

Recall that we control the size of #abs(x-(-1.5)) = abs(x-(-3/2))#

I see #abs(x+3# which is the same as #abs(x-(-3))# so let's 'factor out #-2#.

# = abs(-2)abs(x+3/2)#

# = 2 abs(x-(-3/2))#

In order to make this less than #epsilon#, it suffices to make #abs(x-(-1.5))# less than #epsi/2#

Proving our L is correct -- Writing the proof

Claim: #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#

Proof:

Given #epsilon > 0#, choose #delta = epsilon/2#. (Note that #delta# is positive.)

Now if < delta# then

#abs((9-4x^2)/(3+2x) -6) = abs(((3-2x)(3+2x))/(3+2x) - 6)#

# = abs(((3-2x) - 6)#

#=abs(-2x-3)#

# = abs(-2)abs(x+3/2)#

# = 2 abs(x-(-3/2))#

# < 2 delta#

# = 2 (epsi/2)#

# = epsilon#

We have shown that for any positive #epsilon#, there is a positive #delta# such that for all #x#, if #0 < abs(x-(-1.5)) < delta#, then #abs((9-4x^2)/(3+2x) -6) < epsilon#.

So, by the definition of limit, we have #lim_(xrarr-1.5)((9-4x^2)/(3+2x) ) = 6#.

Note

This is an exampole of a limit in which the strict inequality #0 < abs(x-delta)# is very important. If we allowed #0 = abs(x-(-1.5))#, then a choice of #x = -1.5# would result in an undefined expression. #abs((9-4x^2)/(3+2x) -6) #.

4

Answer:

# 4sqrt(3)pia^3 #

Explanation:

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

enter image source here

The red circle has radius #2a#, hence its equation is:

# x^2+y^2=(2a)^2 => x^2+y^2=4a^2#

Method 1 - Calculate core and subtract from Sphere

First let us consider the volume of the entire Sphere, which has radius #2a#. I will use the standard volume formula #V=4/3pir^3#

# :. V_("sphere") = 4/3 pi (2a)^3 #
# " " = 4/3 pi 8a^3 #
# " " = 32/3 pi a^3 #

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about #Oy#, the shell formula is:

# V = 2pi \ int_(alpha)^(beta) \ xf(x) \ dx #

Also note that we require twice the volume because we have a portion above and below the #x#-axis. The shell volume of revolution about #Oy# for the bore is given by:

# V_("bore") = 2pi \ int_(0)^(a) \ x(2sqrt(4a^2-x^2)) \ dx #
# " " = 2pi \ int_(0)^(a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let #u=4a^2-x^2 => (du)/dx = -2x#
When # { (x=0), (x=a) :} => { (u=4a^2), (u=3a^2) :}#

And so:

# V_("bore") = -2pi \ int_(4a^2)^(3a^2) \sqrt(u) \ du #
# " " = 2pi \ int_(3a^2)^(4a^2) \sqrt(u) \ du #
# " " = 2pi \ [ u^(3/2)/(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(3a^2)^(4a^2) #
# " " = (4pi)/3 \ ((4a^2)^(3/2) - (3a^2)^(3/2)) #
# " " = 4/3pi \ (8a^3-3sqrt(3)a^3) #
# " " = 32/3pi a^3-4sqrt(3)pia^3 #

And so the total volume is given by:

# V_("total") = V_("sphere") - V_("bore" #
# " " = 32/3 pi a^3 - (32/3pi a^3-4sqrt(3)pia^3)#
# " " = 4sqrt(3)pia^3#

Method 2 - Calculate volume of bead directly

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about #Oy#, but this time we will consider the bead itself rather than the core that is removed, again we need to twice the volume because we have a portion above and below the #x#-axis. The shell volume of revolution about #Oy# for the bead is given by:

# V_("total") = 2pi \ int_(a)^(2a) \ x(2sqrt(4a^2-x^2)) \ dx #
# " " = 2pi \ int_(a)^(2a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let #u=4a^2-x^2 => (du)/dx = -2x#
When # { (x=a), (x=2a) :} => { (u=3a^2), (u=0) :}#

And so:

# V_("total") = -2pi \ int_(3a^2)^(0) \sqrt(u) \ du #
# " " = 2pi \ int_(0)^(3a^2) \sqrt(u) \ du #
# " " = 2pi \ [ u^(3/2)/(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(0)^(3a^2) #
# " " = (4pi)/3 \ ((3a^2)^(3/2) - 0) #
# " " = (4pi)/3 \ (3sqrt(3)a^3 )#
# " " = 4sqrt(3)pia^3 #, as above

2

Answer:

See below.

Explanation:

Calling #f(x) = (1/2+x)(1/3-x) = -x^2/6-x/6+1/6# and

we know that the local minima/maxima are given at the stationary points or when

#d/(dx)f(x) = 0#

but in our case we have

#f(g(x))# so the stationary points are when

#d/(dx)f(g(x))=-1/6(1+12 g(x))g'(x)=0#

with #g(x) = cos(x)#

so the stationary points are the solutions of

#{(1+12g(x)=1+12cos(x)=0),(g'(x)=-sin(x)=0):}#

and the stationary points are

#x_1 = -"arccos"(-1/12)+2kpi# and

#x_2 = k pi# for #k in ZZ#

The stationary points qualification is done using #d^2/(dx^2)f(g(x))#

or

#d^2/(dx^2)f(g(x))=-2g'(x)^2 - 1/6 (1 + 12 g(x)) g''(x)#

or

#d^2/(dx^2)f(cos(x))=Cos(x)/6 + 2 Cos^2(x) - 2 Sin^2(x)#

Regarding the stationary points the value of #d^2/(dx^2)f(cos(x))#qualifies the type of extremum. If #d^2/(dx^2)f(cos(x)) > 0 # is a local minimum and if #d^2/(dx^2)f(cos(x)) < 0# is a local maximum.

so

#d^2/(dx^2)f(cos(x_1)) = -143/72# local maxima
#d^2/(dx^2)f(cos(x_2)) = 13/6# local minima

Attached a plot showing this behavior.

enter image source here

so concluding, #x_1# for the maxima and #x_2# for the minima

NOTE. The same result for the maximum point would be obtained directly by making

#(df)/(dx)=-1/6 - 2 x=0->x=-1/12# with the maximum value given by

#f(-1/12)=25/144# and here

#(d^2f)/(dx^2)(-1/12)=-2# qualifying a maximum.

3

Answer:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x sinh^(-1) x - sqrt(x^2+1) + C#

#color(white)(int ln(x+sqrt(x^2+1)) dx) = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C#

Explanation:

Let us try a hyperbolic substitution.

Let #x = sinh theta#

Then:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = int color(white)(.)ln(sinh theta + sqrt(sinh^2 theta + 1)) dx/(d theta) d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(sinh theta + cosh theta) cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(e^theta) cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)theta cosh theta color(white)(.)d theta#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = theta sinh theta - cosh theta + C#

#color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = x sinh^(-1) x - sqrt(x^2+1) + C#

If we would prefer not to have an inverse hyperbolic function in the answer, let's find another expression for it:

Let:

#y = sinh^(-1) x#

Then:

#x = sinh y = 1/2(e^y-e^(-y))#

Hence:

#e^y-2x-e^(-y) = 0#

So:

#(e^y)^2-(2x)(e^y)-1 = 0#

So using the quadratic formula:

#e^y = (2x+-sqrt((2x)^2+4))/2#

#color(white)(e^y) = x+-sqrt(x^2+1)#

If #y# is real then #e^y > 0# and we need the #+# sign here.

So:

#e^y = x+sqrt(x^2+1)#

and:

#y = ln(x+sqrt(x^2+1))#

That is:

#sinh^(-1) x = ln(x+sqrt(x^2+1))#

So:

#int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C#

1

Step #1#: Determine the first derivative

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

Step #2#: Determine the critical numbers

These will occur when the derivative equals #0#.

#0 = 6x^2 - 24x + 18#

#0 = 6(x^2 - 4x + 3)#

#0 = (x - 3)(x - 1)#

#x = 3 or 1#

Step #3#: Determine the intervals of increase/decrease

We select test points.

Test point 1: #x = 0#

#f'(0) = 6(0)^2 - 24(0) + 18 = 18#

Since this is positive, the function is uniformly increasing on #(-oo, 1)#.

Test point 2: #x = 2#

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

Since this is negative, the function is decreasing on #(1, 3)#.

I won't select a test point for #(3, oo)# because I know the function is increasing on the interval. The point #x = 3# is referred to as a turning point because it goes from decreasing to increasing or vice versa.

Step #4#: Determine the second derivative

This is the derivative of the first derivative.

#f''(x) = 12x - 24#

Step #5#: Determine the points of inflection

These will occur when #f''(x) = 0#.

#0 = 12x- 24#

#0 = 12(x - 2)#

#x = 2#

Step #6#: Determine the intervals of concavity

Once again, we select test points.

Test point #1#: #x = 1#

#f''(1) = 12(1) - 24 = -12#

This means that #f(x)# concave down (Since it's negative) on#(-oo, 2)#.

This also means that #f(x)# is concave up on #(2, oo)#.

Step #7#: Determine the x/y- intercept

#f(x)# doesn't have any rational factors, so we'll forget about the x-intercepts (as a result, you would need Newton's Method or a similar method to find the x-intercepts).

The y-intercept is

#f(0) = 2(0)^3 - 12(0)^2 + 18x - 1 = -1#

If you can't connect the graph, you could always make a table of values. In the end, you should get a graph similar to the following. graph{2x^3- 12x^2 + 18x - 1 [-10, 10, -5, 5]}

Hopefully this helps!

2

Answer:

For the #2^(nd)# Proof, refer to the Explanation.

Explanation:

Here is a Second Method to prove the Result for

#n>=2, n in NN.#

#u_n=intsin(nx)/sinxdx.#

Now, #sin(nx)/sinx=1/sinx{sin((nx-2x)+2x)}#

Knowing that, #sin(A+B)=sinAcosB+cosAsinB,# we get,

#sin(nx)/sinx=1/sinx{sin(nx-2x)cos2x+cos(nx-2x)sin2x}#

#=1/sinx{(sin((n-2)x))(1-2sin^2x)+(cos((n-2)x))(2sinxcosx)}#

#=1/sinx{sin((n-2)x)-2sin^2xsin((n-2)x)+2sinxcosxcos((n-2)x)}#

#=sin((n-2)x)/sinx-2sinxsin((n-2)x)+2cosxcos((n-2)x)#

#=sin((n-2)x)/sinx+2{cosxcos((n-2)x)-sinxsin((n-2)x)}#

#=sin((n-2)x)/sinx+2{cos((n-2)x+x)}#

#:. sin(nx)/sinx=sin((n-2)x)/sinx+2cos((n-1)x).#

#rArr u_n=intsin(nx)/sinxdx=int{sin((n-2)x)/sinx+2cos((n-1)x)}dx#

#=intsin((n-2)x)/sinxdx+2intcos((n-1)x)dx.#

#"Hence, "u_n=u_(n-2)+(2sin((n-1)x))/(n-1)+C; n in NN, n>=2.#

Rest of the Soln. is the same as in the First Method.

Enjoy Maths.!

1

Answer:

A height of #10# inches would maximize the volume. See below for details.

Explanation:

Let #l# be the length, #w# be the width and #h# be the height. However, since #l = w#, we can narrow ourselves down to two variables, let them be #l# and #h#.

The formula for surface area of a rectangular prism is generally:

#S.A = 2lh + 2lw + 2hw#

In this case, it becomes

#600 = 2lh+ 2l^2 + 2lh#

#600 = 4lh + 2l^2#

We want to solve for one of the variables now.

#600 - 2l^2 = 4lh#

#(600 -2l^2)/(4l) =h#

#(300 - l^2)/(2l) = h#

The volume of a rectangular prism is given by:

#V =l xx w xx h#

Which in our case becomes:

#V = l^2 xx h#

Now:

#V = l^2(300 - l^2)/(2l)#

#V = 1/2l(300 - l^2)#

#V = 1/2(300l - l^3)#

#V = 150l - 1/2l^3#

Now find the derivative of this with respect to volume.

#V' = 150 - 3/2l^2#

Find the critical values. These will occur when the derivative equals #0#.

#0 = 150 - 3/2l^2#

#3/2l^2 = 150#

#l^2 = 100#

#l = +- 10#

A negative side length doesn't make sense, so we don't accept it as a solution. We now verify graphically to insure this is a maximum.

enter image source here

And it is!

We realize the y-coordinate of the maximum is #y =1000#, which is the maximum volume possible. All that is left for us to do is find the height that produces the maximum volume.

#V = l^2 * h#

#1000 = (10)^2 * h#

#1000 = 100h#

#h = 10#

Hopefully this helps!