To find the critical points of a function, first ensure that the function is differentiable, and then take the derivative. Next, find all values of the function’s independent variable for which the derivative is equal to 0, along with those for which the derivative does not exist. These are our critical points.
The critical points of a function ##f(x)## are those where the following conditions apply:
A) The function exists.
B) The derivative of the function ##f'(x)## is either equal to 0 or does not exist.
As an example with a polynomial function, suppose I take the function ##f(x) = x^2 + 5x – 7## The derivative of this function, according to the power rule, is the function ##f'(x) = 2*x + 5##.
For our first type of critical point, those where the derivative is equal to zero, I simply set the derivative equal to 0. Doing this, I find that the only point where the derivative is 0 is at ##x = -2.5##, at which value ##f(x) = -13.25##.
For our second type of critical point, I look to see if there are any values of ##x## for which my derivative does not exist. I see there are none, so I am confident in stating that the only critical point on my function occurs at ##(-2.5, -13.25)##
For a slightly more tricky example, we will take the function ##f(x) = x^(2/3)## Differentiation yields ##f'(x) = (2/3)*x^(-1/3)## or ##f'(x) = 2/(3x^(1/3))##. In this example, there are no real numbers for which ##f'(x)=0##, but there is one where ##f'(x)## does not exist, namely at ##x=0##. The original function, however, does exist at this point, thus satisfying condition A from the summary. Therefore, this function possesses a critical point at ##(0, 0)##.
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How Do You Find F O G X And Its Domain G O F X And Its Domain F O G 2 And G O F
/in Uncategorized /by developerGiven##color(white)(“XXX”)f(color(blue)(x))=color(blue)(x)^2-1##and##color(white)(“XXX”)g(color(red)(x))=color(red)(x)+1##
Note that ##(f@g)(x)## can be written ##f(g(x))## and that ##(g@f)(x)## can be written ##g(f(x))##
##(f@g)(x) = f(color(blue)(g(x))) = color(blue)(g(x))^2-1####color(white)(“XXXXXX”)=(color(blue)(x+1))^2-1####color(white)(“XXXXXX”)=x^2+2x##Since this is defined for all Real values of ##x##,the of ##(f@g)(x)## is all Real values.(although it wasn’t asked for, the would be ##[-1,+oo)##)
Similarly##(g@f)(x)=g(color(red)(f(x)))+1####color(white)(“XXXXXX”)=g(color(red)(x^2-1))####color(white)(“XXXXXX”)=color(red)(x^2-1)+1####color(white)(“XXXXXX”)=x^2##Again, this is defined for all Real values of ##x##so the Domain is all Real values.(but the Range is ##[0,+oo)##)
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How Do You Find The Area Of Circle X 2 Y 2 5 625
/in Uncategorized /by developerCartesian equation of a circle: ##x^2+y^2=r^2##Therefore ##r^2 = 5625## units.Area of a circle : ##A=pi.r^2=5625pi## units
No explanation required …
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How Do You Find The Center Radius Form Of The Equation Of The Circle Described A
/in Uncategorized /by developerUse the equation and plug in the center and the radius.
##(x+2)^2+y^2=25##
Recall the equation for a circle:
##(x-h)^2+(y-k)^2=r^2##
Where ##(h,k)## is the center of a circle with radius ##r##.
Therefore, a circle with radius ##5##, centered at ##(-2,0)## has the equation:
##color(red)((x+2)^2+y^2=25##
And the graph would look like
graph{(x+2)^2+y^2=25 [-14.24, 14.24, -7.12, 7.12]}
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How Do You Find The Critical Points Of A Rational Function
/in Uncategorized /by developerTo find the critical points of a function, first ensure that the function is differentiable, and then take the derivative. Next, find all values of the function’s independent variable for which the derivative is equal to 0, along with those for which the derivative does not exist. These are our critical points.
The critical points of a function ##f(x)## are those where the following conditions apply:
A) The function exists.
B) The derivative of the function ##f'(x)## is either equal to 0 or does not exist.
As an example with a polynomial function, suppose I take the function ##f(x) = x^2 + 5x – 7## The derivative of this function, according to the power rule, is the function ##f'(x) = 2*x + 5##.
For our first type of critical point, those where the derivative is equal to zero, I simply set the derivative equal to 0. Doing this, I find that the only point where the derivative is 0 is at ##x = -2.5##, at which value ##f(x) = -13.25##.
For our second type of critical point, I look to see if there are any values of ##x## for which my derivative does not exist. I see there are none, so I am confident in stating that the only critical point on my function occurs at ##(-2.5, -13.25)##
For a slightly more tricky example, we will take the function ##f(x) = x^(2/3)## Differentiation yields ##f'(x) = (2/3)*x^(-1/3)## or ##f'(x) = 2/(3x^(1/3))##. In this example, there are no real numbers for which ##f'(x)=0##, but there is one where ##f'(x)## does not exist, namely at ##x=0##. The original function, however, does exist at this point, thus satisfying condition A from the summary. Therefore, this function possesses a critical point at ##(0, 0)##.
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How Do You Find The Derivative Of Cos 2 2x
/in Uncategorized /by developer##y=cos^2(2x)####dy/dx=2cos(2x)*-sin(2x)*2####dy/dx=-4cos(2x)*sin(2x)####dy/dx=-2sin(4x)##
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How Do You Find The Derivative Of Cos Pi X
/in Uncategorized /by developerThe derviative is ##-pisin(pix)##
You must use the chain rule to calculate this derivative since there is a function (##pi*x##) inside of another function (the cosine function ##cos(pix)##).
First, take the derivative of the outside function (which is cosine in this case) without touching the inside and then multiply by the derivative of the inside function (##pix## for this problem).
The derivative of ##cos(x)## is ##-sin(x)## and the derivative of ##pi*x## is ##pi## (since ##pi## is a constant, albeit an irrational one) so the derivative for the entire thing is ##-sin (pi*x) * (pi)##.
Hope that helps!
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How Do You Find The Derivative Of Cotx
/in Uncategorized /by developer##dy/dx = -csc^2x##
##y = cotx##
##y = 1/tanx##
##y = 1/(sinx/cosx)##
##y = cosx/sinx##
Letting ##y= (g(x))/(h(x))##, we have that ##g(x) = cosx## and ##h(x) = sinx##.
##y’ = (g'(x) xx h(x) – g(x) xx h'(x))/(h(x))^2##
##y’ = (-sinx xx sinx – (cosx xx cosx))/(sinx)^2##
##y’ = (-sin^2x – cos^2x)/(sinx)^2##
##y’ = (-(sin^2x + cos^2x))/sin^2x##
##y’ = -1/sin^2x##
##y’ = -csc^2x##
Hopefully this helps!
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How Do You Find The Derivative Of Y Arcsin 2x 1
/in Uncategorized /by developerIn these cases you need chain rule .
##dy/(dx)## =##d/(dz## x##dz/dx##===>
It’s better to take ##d/(dz## & ## dz/dx## separately .
##d/(dz##= ##d/(d(2x+1)##.##arcsin (2x+1)##.
##d/(dz##=##1/sqrt(1-(2x+1)^2##= ##1/(2sqrt(-x^2-x)## ——(1)
##dz/dx##= ##d/dx## .## (2x+1)## =##d/dx 2x## = ##2## ——(2)
Then multiply (1) &(2),
So finally you get ##dy/(dx)##= ##1/(2sqrt(-x^2-x)####2##
##dy/(dx)##=##1/(sqrt(-x^2-x)## ; where ##y## = ##arcsin(2x+1)##
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How Do You Find The Equation Of A Parabola With Vertex At The Origin And Directr
/in Uncategorized /by developer##y^2=8x##
The standard form of the parabola is ##y^2=4ax##, giving a parabola with its axis parallel to the ##x##-axis, vertex at the origin, focus ##(a,0)## and directrix ##x=-a##. So in your case ##a=2##, giving ##y^2=4ax##.
Alternatively, you can from a definition of a parabola, which is the set of all points ##(x,y)## such that the distance from the point to the directrix ##x=-2## is the same as the distance to the focus (2,0).(The vertex is half-way between the focus and the directrix.)
##(x-(-2))^2=(x-a)^2+y^2####cancel(x^2)+4ax+cancel 4=cancel(x^2)-4ax+cancel 4+y^2####y^2=4ax+4ax=8ax##
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How Do You Find The Equation Of The Plane In Xyz Space Through The Point P 4 5 4
/in Uncategorized /by developerThe answer is: ##5x+3y+4z-51=0##
Given a poiint ##P(x_p,y_p,z_p)## and a vector ##vecv(a,b,c)## perpendicular to the plane, the equation is:
##a(x-x_p)+b(y-y_p)+c(z-z_p)=0##
So:
##-5(x-4)-3(y-5)-4(z-4)=0rArr##
##-5x+20-3y+15-4z+16=0rArr##
##5x+3y+4z-51=0##
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