How do you find the maximum, minimum and inflection points and concavity for the function #g(x) = 170 + 8x^3 + x^4#?

Question

2968 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T17:32:02

Point of minima #x=-6#, points of inflection #x=0, x=-4# & the function is concave in #x\in(-6, 0)#

The given function:

#g(x)=170+8x^3+x^4#

#g'(x)=24x^2+4x^3#

#g''(x)=48x+12x^2#

For mamima or minima we have

#g'(x)=0#

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

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

#x=0, -6 #

Now, #g''(0)=48(0)+12(0)^2=0#

hence, #x=0# is a point of inflection

Now, #g''(-6)=48(-6)+12(-6)^2#

#=144>0#

hence, #x=-6# is a point of maxima

Now, for points of inflection we have

#g''(x)=0#

#48x+12x^2=0#

#12x(4+x)=0#

#x=0, -4#

The curve will be concave iff

#g'(x)<0#

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

#4x^2(6+x)<0#

#x\in(-6, 0)#

Thus, the graph will be concave for #x\in(-6, 0)#