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Answer:

Please refer to the explanation below.

Explanation:

We have to take the derivative of the function #f(x)#, and then plug in #x=2# into #f'(x)#.

#f(x)=(x^2-2)/(x-1)#

So, we have to use the quotient rule, which states that,

#d/dx(u/v)=(u'v-uv')/v^2#

Here, #u=x^2-2#, #v=x-1#

#:.f'(x)=(2x(x-1)-1*(x^2-2))/(x-1)^2#

#f'(x)=(2x^2-2x-x^2+2)/(x-1)^2#

#f'(x)=(x^2-2x+2)/(x-1)^2#

Plugging in #x=2#, we get

#f'(2)=(2^2-2*2+2)/(2-1)^2#

#f'(2)=(4-4+2)/1^2#

#f'(2)=2/1#

#f'(2)=2#

So, the instantaneous rate of change will be #2#.

Answer:

#(2,8)#

Explanation:

Two lines are parallel if they share the same slope.

Any line parallel to #y=10x# will have the slope 10.

Now, we have: #y=3x^2-2x#

We try to find #dy/dx#

=>#d/dx(y)=d/dx(3x^2-2x)#

We use the power rule:

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.

#dy/dx=3*2x^(2-1)-2*1x^(1-1)#

=>#dy/dx=6x-2#

Now, we set #dy/dx=10#

=>#10=6x-2#

=>#12=6x#

=>#2=x#

We plug this value in to get our #y#, or #f(x)#.

#3(2)^2-2(2)=y#

=>#12-4=y#

=>#8=y#

Therefore, the point is at #(2,8)#

Answer:

Use the substitution #sqrtx+1=sqrt7sectheta#.

Explanation:

Let

#I=int1/sqrt(4x+8sqrtx-24)dx#

Complete the square in the square root:

#I=1/2int1/sqrt((sqrtx+1)^2-7)dx#

Apply the substitution #sqrtx+1=sqrt7u#:

#I=int(sqrt7u-1)/sqrt(u^2-1)du#

Rearrange:

#I=sqrt7intu/sqrt(u^2-1)du-int1/sqrt(u^2-1)du#

Apply the substitution #u=sectheta#:

#I=sqrt7sqrt(u^2-1)-intsecthetad theta#

Hence

#I=sqrt(7u^2-7)-ln|sectheta+tantheta|+C#

Reverse the last substitution:

#I=sqrt(7u^2-7)-ln|u+sqrt(u^2-1)|+C#

Reverse the first substitution:

#I=sqrt((sqrtx+1)^2-7)-ln|(sqrtx+1)+sqrt((sqrtx+1)^2-7)|+C#

Simplify:

#I=sqrt(x^2+2sqrtx-6)-ln|(sqrtx+1)+sqrt(x^2+2sqrtx-6)|+C#

Answer:

Integral is #0#.

Explanation:

Let #x-3=t# then #dx=dt# and

#int_2^4(x-3)^3dx=int_-1^1t^3dt#

= #[t^4/4]_-1^1#

= #[1/4-1/4]=0#

Note that when #x=2#, #t=-1# and when #x=4#, #t=1#.

Now observe the graph of #(x-3)^3#, as given below

graph{(x-3)^3 [0.542, 5.542, -1.2, 1.3]}

Note that area of curve in the interval #[2,3]# is symmetric w.r.t. area of curve in the interval #[3,4]#, but they are in opposite direction, Hence the integral is #0#.

Answer:

See below.

Explanation:

The critical points are the points where the first derivative equals zero, or the point doesn’t exist.

This is a polynomial so it is continuous for all #x in RR#

First derivative of #f(x)=x^4-8x^3+18x^2-5#

#dy/dx(f(x))=4x^3-24x^2+36x#

Equating this to zero:

#4x^3-24x^2+36x=0#

#4x(x^2-6x+9)=0#

#4x=0=>x=0#

#(x^2-6x+9)=0#

Factor:

#(x-3)^2=0=>x=3#

Putting these values in #f(x)#

#f(0)=-5#

#f(3)=22#

Critical Points:

#color(blue)(x=0)# and #color(blue)(x=3)#

#(0,-5)# and #(3,22)#

#f(x)# is increasing where #f'(x)>0#

#f(x)# is decreasing where #f'(x)<0#

#4x^3-24x^2+36x>0#

#x^3-6x^2+9x>0#

#x(x^2-6x+9)>0#

#x(x-3)^2>0#

#0 < x < 3# , #3< x < oo#

Increasing in #(0 , 3)# and #(3,oo)#

#4x^3-24x^2+36x<0#

#x^3-6x^2+9x<0#

#x(x^2-6x+9)<0#

#x(x-3)^2<0#

#-oo < x < 0#

Decreasing in #(-oo,0)#

GRAPH:

enter image source here

Answer:

#lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2#

Explanation:

We have

#lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = lim_(n->oo)sum_(k=1)^(n) 1/sqrt(2n^2+k) =#

#= lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)#

Now, let's try to remember the definition of an integral. This might seem strange, but it will come of great use.

The definite integral #int_a^b f(x)dx# represents the #color(red)("net area")# defined by the graph of the function #f(x)# and the x-axis. One way to find the value of an integral is by aproximating it with infinitely many rectangles ("boxes") of equal or not width.

This concept is called a #color(red)("Riemann Sum")#.

There are infinitely many rectangles, therefore, if there are #n# rectangles, we have to take the limit as #n->oo#.

Now, let's take the case where the width is constant between all boxes. If we define #x_1,x_2,...,x_n# to be some "marks" on the x-axis such that the width of the #i#-th rectangle is #x_(i+1) - x_i = Delta x_i#, then its lenght is #f(x_i)#. Since the width is equal, #Deltax_i# is the same for all #i#'s. We can more easily call it #color(red)(Delta#.

The Riemann Sum aproximates an integral by being the sum of all the rectangles. In order to simplify this further, let's take the particular case where #color(red)(a = 0# and #color(red)(b=1#. We have:

#int_0^1 f(x)dx = lim_(n->oo) sum_(i=1)^(n) Delta*f(x_i)#

This is only true if #x_n# is equal to #1#.

One way to make #Delta# constant is to define #x_k = k/n#, for some #k#, #1<=k<=n#. This way, #x_n = 1# and

#color(red)(Delta) = x_(i+1)-x_i = (i+1)/n-i/n = color(red)(1/n)#.

#int_0^1 f(x)dx = lim_(n->oo) sum_(i=1)^n 1/n * f(i/n)#.

This looks familiar to our original sum, #lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)#.

All we are left to do is to assume they're equal and then solve the integral.

#lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2) = lim_(n->oo) sum_(i=1)^n 1/n * f(i/n)#

#i# and #k# are just the indexes, the name of which doesn't matter. Both sums could be counting #i# or #k#, it's irrelevant.

#lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2) = lim_(n->oo) sum_(k=1)^n 1/n * f(k/n)#

#=> f(k/n) = 1/sqrt(2+k/n^2) => f(x) = 1/sqrt(2+x/n)#.

We have to find the integral of this function:

#lim_(n->oo) int_0^1 dx/sqrt(2+x/n)#

Since #n->oo#, this means that #x/n = 0#, as #x# is finite.

#lim_(n->oo) int_0^1 dx/sqrt2 = 1/sqrt2#.

This is because it forms a rectangle with the x-axis, the width and lenght of which is #1# and #1/sqrt2#, respectively.

As a conclusion,

#color(blue)(lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2#.

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  • Sonnhard answered · 5 hours ago
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