Exam 3: 3.1 Concavity

Definition 3.1. A graph is called concave up on an interval (a, b) if the graph looks like:

A graph is called concave down on an interval (a, b) if the graph looks like:

Remark 3.2 (Finding inflection points). For a function y=f(x)y=f(x). The point p is a possible inflection point if f(p)(2)=0f\left(p\right)^{\left(2\right)}=0 orf(2)(p)f^{\left(2\right)}\left(p\right) undefined. To test whether p is actually an inflection point, check whether f(2)f^{\left(2\right)} changes sign at p (using a sign chart).

Example 3.3. Find the inflection points for f(x)=14x4−4x3+24x2f\left(x\right)=\frac14x^4-4x^3+24x^2

Step 1: Solve f(2)(x)=0f^{\left(2\right)}\left(x\right)=0 or und

f′(x)=x3−12x2+48xf^{\prime}\left(x\right)=x^3-12x^2+48x

f(2)(x)=3x2−24x+483=03f^{\left(2\right)}\left(x\right)=\frac{3x^2-24x+48}{3}=\frac03 (Divide both sides by 3)

f(2)(x)=x2−8x+16=0f^{\left(2\right)}\left(x\right)=x^2-8x+16=0

f(2)(x)=(x−4)(x−4)=0f^{\left(2\right)}\left(x\right)=\left(x-4\right)\left(x-4\right)=0

f(2)(x)=x−4=0f^{\left(2\right)}\left(x\right)=x-4=0

f(2)(x)=x=4f^{\left(2\right)}\left(x\right)=x=4 (possible inflection point)

Step 2: Determine which of the possibilities is an IP

f(2)(0)=48f^{\left(2\right)}\left(0\right)=48

f(2)(5)=3f^{\left(2\right)}\left(5\right)=3

x=4x=4 is not an inflection point

f(x) has no IP

Example 3.4. Find the inflection points for f(x)=xe−xf\left(x\right)=xe^{-x}

f′(x)=(1)(e−x)+(x)(e−x(−1))f^{\prime}\left(x\right)=\left(1\right)\left(e^{-x}\right)+\left(x\right)\left(e^{-x}\left(-1\right)\right)

f′(x)=e−x−xe−xf^{\prime}\left(x\right)=e^{-x}-xe^{-x}

f′(x)=(1−x)e−1f^{\prime}\left(x\right)=\left(1-x\right)e^{-1}

f(2)(x)=(−1)(e−x)+(1−x)(e−x(−1))f^{\left(2\right)}\left(x\right)=\left(-1\right)\left(e^{-x}\right)+\left(1-x\right)\left(e^{-x}\left(-1\right)\right)

f(2)(x)=−e−x−(1−x)e−xf^{\left(2\right)}\left(x\right)=-e^{-x}-\left(1-x\right)e^{-x}

f(2)(x)=−e−x−(e−x−xe−x)f^{\left(2\right)}\left(x\right)=-e^{-x}-\left(e^{-x}-xe^{-x}\right)

f(2)(x)=−e−x−e−x+xe−xf^{\left(2\right)}\left(x\right)=-e^{-x}-e^{-x}+xe^{-x}

f(2)(x)=2e−x+xe−xf^{\left(2\right)}\left(x\right)=2e^{-x}+xe^{-x}

f(2)(x)=(−2+x)e−x=0f^{\left(2\right)}\left(x\right)=\left(-2+x\right)e^{-x}=0

−2+x=0-2+x=0

x=2x=2

e−x=0e^{-x}=0

No solution

(−∞,2)\left(-\infty,2\right) Concave down

(2,∞)\left(2,\infty\right) Concave up

We gave already seen the first-derivative test for local extrema, i.e.

How can we use concavity and the second derivative to test for local extrema?

Theorem 3.5 (second-derivative test). Let f(x)f(x) be a differentiable function.

  • If f′(p)=0f^{\prime}\left(p\right)=0 andf^{\left(2\right)}\left(p\right)>0 , then f(x)f(x) has a local minimum at p.

    • Concave up

  • f(2)(x)=(−2x)(x2+1)2−4x(1−x2)(x2+1)(x2−1)4f^{\left(2\right)}\left(x\right)=\frac{\left(-2x\right)\left(x^2+1\right)^2-4x\left(1-x^2\right)\left(x^2+1\right)}{\left(x^2-1\right)^4} I

  • f f′(p)=0f^{\prime}\left(p\right)=0 andf^{\left(2\right)}\left(p\right)<0 , then f(x)f(x) has a local maximum at p.

    • Concave down

  • If f′(p)=0f^{\prime}\left(p\right)=0 andf(2)(x)=0f^{\left(2\right)}\left(x\right)=0 , then the test doesn’t tell us anything.

    • Go back and try the first derivative test.

Example 3.6. Determine the local extrema for f(x)=xx2+1f\left(x\right)=\frac{x}{x^2+1}

f′(x)=(1)(x2+1)−(x)(2x)(x2+1)2f^{\prime}\left(x\right)=\frac{\left(1\right)\left(x^2+1\right)-\left(x\right)\left(2x\right)}{\left(x^2+1\right)^2}

f′(x)=(x2+1)−2x2(x2+1)2f^{\prime}\left(x\right)=\frac{\left(x^2+1\right)-2x^2}{\left(x^2+1\right)^2}

f′(x)=1−x2(x2+1)2f^{\prime}\left(x\right)=\frac{1-x^2}{\left(x^2+1\right)^2}

f′(x)=(x2+1)21−x2(x2+1)2=0(x2+1)2f^{\prime}\left(x\right)=\left(x^2+1\right)^2\frac{1-x^2}{\left(x^2+1\right)^2}=0\left(x^2+1\right)^2 (Multiply by (x²+1)² on both sides)

f′(x)=1−x=0f^{\prime}\left(x\right)=1-x=0

f′(x)=1=x2f^{\prime}\left(x\right)=1=x^2

f′(x)=±1=xf^{\prime}\left(x\right)=\pm1=x

f(2)(x)=(−2x)(x2+1)2−(1−x2)(2(x2+1)2⋅2x)[(x2−1)2]2f^{\left(2\right)}\left(x\right)=\frac{\left(-2x\right)\left(x^2+1\right)^2-\left(1-x^2\right)\left(2\left(x^2+1\right)^2\cdot2x\right)}{\left\lbrack\left(x^2-1\right)^2\right\rbrack^2}

f(2)(x)=(−2x)(x−1)−4x(x2+1)(1−x2)(x2−1)4f^{\left(2\right)}\left(x\right)=\frac{\left(-2x\right)\left(x-1\right)-4x\left(x^2+1\right)\left(1-x^2\right)}{\left(x^{^2}-1\right)^4}

f(2)(x)=(x−1)[−2x(x2+1)−4x(1−x2)](x2+1)3f^{\left(2\right)}\left(x\right)=\frac{\left(x-1\right)\left\lbrack-2x\left(x^2+1\right)-4x\left(1-x^2\right)\right\rbrack}{\left(x^2+1\right)^3}

f(2)(x)=−2x(x2+1)−4x(1−x2)(x2+1)3f^{\left(2\right)}\left(x\right)=\frac{-2x\left(x^2+1\right)-4x\left(1-x^2\right)}{\left(x^2+1\right)^3}

f^{\left(2\right)}\left(1\right)=\frac{\left(-2\right)\left(2\right)-\left(4\right)\left(0\right)}{2^3}=\frac{-4}{8}<0 (max @ x=1)

f^{\left(2\right)}\left(-1\right)=\frac{\left(2\right)\left(2\right)+4\left(0\right)}{2^3}=\frac48>0 (min @ x=-1