Find points of inflection and discuss concavity

Reference no: EM13715514

Find points of inflection and discuss concavity for f(x)=x(x-4)3+k.

Use k=2

example provided by teacher: SOLUTION:
Let's use the product and chain rule to find f'(x)= x*3(x-4)^2 + (x-4)3; factor out the common binomial to get f'(x)=[3x+x-4](x-4)2= [4x-4](x-4)2= 4[x-1](x-4)2; Now to f'' let's use the product and chain rule again. f''(x)=4[x-1]*2(x-4) + 4(x-4)2; factor out the 4(x-4) to get f''(x)=4(x-4){2x-2+x-4}=4(x-4){3x-6}=12(x-4){x-2}. Setting this equal to zero gives the point of inflection as x=2 and x=4. Let's take f''(3) to see the concavity there.

SOLUTION: f''(3)<0 so f(x) has a negative concavity and is concave down. f(x) will be concave UP on (-infinity,2) and (4,+infinity).

If you can, it is a GREAT IDEA to graph this function to see if the concavity and points of inflection are consistent with our findings.

Reference no: EM13715514

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Find points of inflection and discuss concavity : Let's use the product and chain rule to find f'(x)= x*3(x-4)^2 + (x-4)3; factor out the common binomial to get f'(x)=[3x+x-4](x-4)2= [4x-4](x-4)2= 4[x-1](x-4)2; Now to f'' let's use the product and chain rule again. f''(x)=4[x-1]*2(x-4) + 4(x-4..

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