Graph f(x)=3x^4+4x^3

Problem

ƒ(x)=3*x4+4*x3

Solution

1. Find the intercepts by setting ƒ(x)=0 for x-intercepts and evaluating ƒ(0) for the y-intercept.

ƒ(x)=x3*(3*x+4)

x(3*x+4)3=0⇒x=0,x=−4/3

ƒ(0)=3*(0)4+4*(0)3=0

2. Find the first derivative to determine critical points and intervals of increase or decrease.

d(ƒ(x))/d(x)=12*x3+12*x2

12*x(x+1)2=0⇒x=0,x=−1

3. Determine the nature of critical points using the first derivative test.

ƒ(x)′<0* on *(−∞,−1)

ƒ(x)′>0* on *(−1,0)∪(0,∞)

ƒ*(−1)=3*(−1)4+4*(−1)3=−1

Relative minimum at *(−1,−1)

4. Find the second derivative to determine concavity and points of inflection.

d2(ƒ(x))/(d(x)2)=36*x2+24*x

12*x(3*x+2)=0⇒x=0,x=−2/3

5. Identify concavity intervals using the second derivative.

ƒ(x)″>0* on *(−∞,−2/3)∪(0,∞)* (concave up)

ƒ(x)″<0* on *(−2/3,0)* (concave down)

ƒ*(−2/3)=3*(−2/3)4+4*(−2/3)3=−16/27

Inflection points at *(−2/3,−16/27)* and *(0,0)

6. Determine end behavior by looking at the leading term 3*x4

(lim_x→∞)(ƒ(x))=∞

(lim_x→−∞)(ƒ(x))=∞

Final Answer

ƒ(x)=3*x4+4*x3

The graph is a quartic curve opening upward with x-intercepts at (0,0) and (−4/3,0) a relative minimum at (−1,−1) and inflection points at (−2/3,−16/27) and (0,0)

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