Determine whether a function is convex or concave using the second-order derivative Calculus

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

f\left(x\right)=2x^{2}-4x+5

Calculate the derivative:

f'\left(x\right)=\left(2x^{2}-4x+5\right)'=4x-4

Calculate the second-order derivative:

f''\left(x\right)=\left(4x-4\right)'=4

For any real number x, f''\left(x\right) is positive.

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

f\left(x\right)=x^{3}+6x^{2}-7x-2

Calculate the derivative:

f'\left(x\right)=\left(x^{3}+6x^{2}-7x-2\right)'=3x^{2}+12x-7

Calculate the second-order derivative:

f''\left(x\right)=\left(3x^{2}+12x-7\right)'=6x+12

Solve the equation f''\left(x\right)=0 :

6x+12=0

The solution of the equation is:

x=-2

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

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

Calculate the derivative:

f'\left(x\right)=\left(x^{2}e^{x}\right)'=\left(x^{2}\right)'e^{x}+x^{2}\left(e^{x}\right)'=2xe^{x}+x^{2}e^{x}=\left(x^{2}+2x\right)e^{x}

Calculate the second-order derivative:

f''\left(x\right)=\left(\left(x^{2}+2x\right)e^{x}\right)'=\left(x^{2}+2x\right)'e^{x}+\left(x^{2}+2x\right)\left(e^{x}\right)'=\left(2x+2\right)e^{x}+\left(x^{2}+2x\right)e^{x}=\left(x^{2}+4x+2\right)e^{x}

Solve the equation f''\left(x\right)=0 :

\left(x^{2}+4x+2\right)e^{x}=0

The solutions of the equation are:

x=-2-\sqrt{2} and x=-2+\sqrt{2}

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

f\left(x\right)=x\ln\left(x\right)

Calculate the derivative:

f'\left(x\right)=\left(x\ln\left(x\right)\right)'=\left(x\right)'\ln\left(x\right)+x\left(\ln\left(x\right)\right)'=\ln\left(x\right)+x\cdot\dfrac{1}{x}=\ln\left(x\right)+1

Calculate the second-order derivative:

f''\left(x\right)=\left(\ln\left(x\right)+1\right)'=\dfrac{1}{x}

Solve the equation f''\left(x\right)=0 :

\dfrac{1}{x}=0

The equation does not have any solutions.

We can conclude that f is convex when x\in\left(0,\infty\right).

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

f\left(x\right)=\dfrac{x-1}{x+1}

Calculate the derivative:

f'\left(x\right)=\left(\dfrac{x-1}{x+1}\right)'=\dfrac{\left(x-1\right)'\left(x+1\right)-\left(x-1\right)\left(x+1\right)'}{\left(x+1\right)^{2}}=\dfrac{1\cdot\left(x+1\right)-\left(x-1\right)\cdot1}{\left(x+1\right)^{2}}=\dfrac{2}{\left(x+1\right)^{2}}

Calculate the second-order derivative:

f''\left(x\right)=\left(\dfrac{2}{\left(x+1\right)^{2}}\right)'=\dfrac{-4\left(x+1\right)}{\left(x+1\right)^{4}}=\dfrac{-4}{\left(x+1\right)^{3}}

The sign of f'' is the following:

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

f\left(x\right)=x^{4}-3x^{2}+7

Calculate the derivative:

f'\left(x\right)=\left(x^{4}-3x^{2}+7\right)'=4x^{3}-6x

Calculate the second-order derivative:

f''\left(x\right)=\left(4x^{3}-6x\right)'=12x^{2}-6

Solve the equation f''\left(x\right)=0 :

12x^{2}-6=0

The solutions of the equation are:

x=-\dfrac{\sqrt{2}}{2} and x=\dfrac{\sqrt{2}}{2}

The sign of f'' is the following:

x	-\infty -\dfrac{\sqrt{2}}{2} \dfrac{\sqrt{2}}{2} \infty
f''\left(x\right)	+ + + + + + + + + + + + + + 0 - - - - - - - - - - - - 0 + + + + + + + + + + + + + + + +

Recall that:

• If f''\left(x\right) \gt 0 when x\in\left( a,b \right), then f is convex on the interval \left( a,b \right).
• If f''\left(x\right) \lt 0 when x\in\left( a,b \right), then f is concave on the interval \left( a,b \right).

Here, we have:

f\left(x\right)=x^{2}\sqrt{x}

Calculate the derivative:

f'\left(x\right)=\left(x^{2}\sqrt{x}\right)'=\left(x^{2}\cdot x^{\frac{1}{2}}\right)'=\left(x^{\frac{5}{2}}\right)'=\dfrac{5}{2}\cdot x^{\frac{5}{2}-1}=\dfrac{5}{2}\cdot x^{\frac{3}{2}}

Calculate the second-order derivative:

f''\left(x\right)=\left(\dfrac{5}{2}\cdot x^{\frac{3}{2}}\right)'=\dfrac{5}{2}\cdot \dfrac{3}{2}\cdot x^{\frac{3}{2}-1}=\dfrac{5}{2}\cdot \dfrac{3}{2}\cdot x^{\frac{1}{2}}=\dfrac{15}{4}\sqrt{x}

Solve the equation f''\left(x\right)=0 :

\dfrac{15}{4}\sqrt{x}=0

The solution of the equation is:

x=0

We conclude that f''\left(x\right) \gt 0 when x\in\left(0,\infty\right), so f is convex when x\in\left(0,\infty\right).