What are the points of inflection of #f(x)=(3x^2 + 8x + 5)/(4-x)
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Question

What are the points of inflection of #f(x)=(3x^2 + 8x + 5)/(4-x)
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Christopher Agresta · Accepted Answer

Even though there are no inflexion points, there still are minimum#(-1.323,-0.063)# and maximum#(9.323,-63.937)#.

Some basic rules of differentiation are as follows, where #u# and #v# are functions of #x#: Addition / subtraction rule If #y=u+-v#, #dy/dx=d/dx(u)+-d/dx(v)# Chain rule #dy/dx=dy/(du)xx(du)/dx# Product rule If #y=uv#, #dy/dx=u(dv)/dx+v(du)/dx# Quotient rule If #y=u/v#, #dy/dx=(v(du)/dx-u(dv)/dx)/v^2# Let's get started, #f(x)=(3x^2 + 8x + 5)/(4-x)# First, differentiate #f(x)#, #f'(x)=([d/dx(3x^2 + 8x + 5)] (4-x)-[d/dx(4-x)](3x^2 + 8x + 5))/(4-x)^2# #color(white)(f'(x))=((6x+8) (4-x)-(-1)(3x^2 + 8x + 5))/(4-x)^2# #color(white)(f'(x))=(-6x^2+16x+32+3x^2 + 8x + 5)/(4-x)^2# #color(white)(f'(x))=-(3x^2-24x-37)/(x-4)^2# Since, it is an stationary point, #d/dx[f(x)]=0#, #-(3x^2-24x-37)/(x-4)^2=0# #-3x^2+24x+37=0# #color(white)(xxxxxx/xxxxxx)x=-1.32291 or 9.32291# Find the #y#-coordinate of the stationary points by substituting #x=-1.32291 or 9.32291# into #f(x)#, #y=-0.063 or -63.937# Hence, the co-ordinates of the stationary points are #(-1.323,-0.063)# and #(9.323,-63.937)#. To find the nature of the stationary points, differentiate #f'(x)#, #f''(x)=-([d/dx (3x^2-24x-37)] (x-4)^2-[d/dx(x-4)^2] (3x^2-24x-37))/(x-4)^4# #color(white)(f''(x))=-((6x-24) (x-4)^2-(2)(x-4)(1)(3x^2-24x-37))/(x-4)^4# #color(white)(f''(x))=-((x-4)[(6x-24)(x-4)-(2)(1)(3x^2-24x-37)])/(x-4)^4# #color(white)(f''(x))=(6x^2-48x+96-6x^2+48x+74)/(x-4)^3# #color(white)(f''(x))=-170/(x-4)^3# Let #x=-1.32291#, #f''(x)=-170/(1.32291-4)^3# #color(white)(f''(x))=8.86054>0# ( minimum ) Let #x=9.32291#, #f''(x)=-170/(9.32291-4)^3# #color(white)(f''(x))=-1.12720<0# ( maximum ) Hence, minimum#(-1.323,-0.063)# and maximum#(9.323,-63.937)#. Check: graph{(3x^2 + 8x + 5)/(4-x) [-320, 320, -160, 160]}

Axel Alvarez · Answer

undefined To find the points of inflection of $ f(x) = \frac{3x^2 + 8x + 5}{4-x} $, first, determine the second derivative, then find where it equals zero, and finally check the concavity to identify the points of inflection.

First derivative:
$ f'(x) = \frac{d}{dx} \left( \frac{3x^2 + 8x + 5}{4-x} \right) $

Second derivative:
$ f''(x) = \frac{d^2}{dx^2} \left( \frac{3x^2 + 8x + 5}{4-x} \right) $

After finding the second derivative, solve for $ x $ when $ f''(x) = 0 $. These values represent possible points of inflection.

Then, determine the concavity of $ f(x) $ around these points to confirm if they are indeed points of inflection. This can be done by evaluating the second derivative's sign around each candidate point.

Once the points of inflection are identified, their corresponding $ y $-coordinates can be found by substituting the $ x $-values into the original function $ f(x) $.

What are the points of inflection of #f(x)=(3x^2 + 8x + 5)/(4-x) #?