Question Statement

Use the second derivative to determine the intervals on which the function is concave upward and concave downward.

(i) $f(x)=-x^2+7x$

(ii) $f(x)=-x^3+6x^2+x-1$

(iii) $f(x)=(x+5{)}^5$

(iv) $f(x)=x(x-4{)}^3$

(v) $f(x)=x^{\frac{1}{2}}+2x$

(vi) $f(x)=x+\frac{9}{x}$

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Background and Explanation

Concavity describes the curvature of a function's graph. A function is concave upward when $f^{\prime \prime }(x)>0$ (shaped like a cup ∪) and concave downward when $f^{\prime \prime }(x)<0$ (shaped like a cap ∩). To find intervals of concavity, compute the second derivative, identify where it equals zero or is undefined (potential inflection points), and test the sign of $f^{\prime \prime }(x)$ in the resulting intervals.

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Solution

Part (i): $f(x)=-x^2+7x$

Given the function: $f(x)=-x^2+7x$

Differentiate with respect to $x$: $f^{\prime }(x)=-2x+7$

Differentiate again with respect to $x$: $f^{\prime \prime }(x)=-2$

Since $f^{\prime \prime }(x)=-2<0$ for all $x\in (-\infty ,\infty )$, the function is concave downward on the entire real line.

Result: Concave downward on $(-\infty ,\infty )$; never concave upward.

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Part (ii): $f(x)=-x^3+6x^2+x-1$

Given the function: $f(x)=-x^3+6x^2+x-1$

Differentiate with respect to $x$: $\begin{array}{rl}{f}^{\prime }(x)& =-3x^2+12x+1\end{array}$

Differentiate again with respect to $x$: $\begin{array}{rl}{f}^{\prime \prime }(x)& =-6x+12\end{array}$

Set $f^{\prime \prime }(x)=0$ to find potential inflection points: $\begin{array}{rl}-6x+12& =0\\ -6x& =-12\\ x& =2\end{array}$

This divides the real line into two intervals: $(-\infty ,2)$ and $(2,\infty )$.

Case I: Consider the interval $(-\infty ,2)$ Take a test point $x=1\in (-\infty ,2)$: $\begin{array}{rl}{f}^{\prime \prime }(1)& =-6(1)+12\\ & =6>0\end{array}$

Since $f^{\prime \prime }(x)>0$ for all $x\in (-\infty ,2)$, the function is concave upward on $(-\infty ,2)$.

Case II: Consider the interval $(2,\infty )$ Take a test point $x=4\in (2,\infty )$: $\begin{array}{rl}{f}^{\prime \prime }(4)& =-6(4)+12\\ & =-24+12\\ & =-12<0\end{array}$

Since $f^{\prime \prime }(x)<0$ for all $x\in (2,\infty )$, the function is concave downward on $(2,\infty )$.

Result: Concave upward on $(-\infty ,2)$; concave downward on $(2,\infty )$.

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Part (iii): $f(x)=(x+5{)}^5$

Given the function: $f(x)=(x+5{)}^5$

Differentiate with respect to $x$ using the chain rule: $\begin{array}{rl}{f}^{\prime }(x)& =5(x+5{)}^4\cdot \frac{d}{dx}(x+5)\\ & =5(x+5{)}^4\cdot (1)\\ f^{\prime }(x)& =5(x+5{)}^4\end{array}$

Differentiate again with respect to $x$: $\begin{array}{rl}{f}^{\prime \prime }(x)& =20(x+5{)}^3\end{array}$

Set $f^{\prime \prime }(x)=0$: $\begin{array}{rl}20(x+5{)}^3& =0\\ (x+5{)}^3& =0\\ x+5& =0\\ x& =-5\end{array}$

The intervals of concavity are $(-\infty ,-5)$ and $(-5,\infty )$.

Case I: Consider the interval $(-\infty ,-5)$ Take $x=-6\in (-\infty ,-5)$: $\begin{array}{rl}{f}^{\prime \prime }(-6)& =20(-6+5{)}^3\\ & =20(-1{)}^3\\ & =-20<0\end{array}$

Since $f^{\prime \prime }(x)<0$, the function is concave downward on $(-\infty ,-5)$.

Case II: Consider $(-5,\infty )$ Take $x=-4\in (-5,\infty )$: $\begin{array}{rl}{f}^{\prime \prime }(-4)& =20(-4+5{)}^3\\ & =20(1{)}^3\\ & =20>0\end{array}$

Since $f^{\prime \prime }(x)>0$, the function is concave upward on $(-5,\infty )$.

Result: Concave downward on $(-\infty ,-5)$; concave upward on $(-5,\infty )$.

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Part (iv): $f(x)=x(x-4{)}^3$

Given: $f(x)=x(x-4{)}^3$

Differentiate with respect to $x$ using the product rule: $\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}(x)\cdot (x-4{)}^3+x\cdot \frac{d}{dx}(x-4{)}^3\\ & =1\cdot (x-4{)}^3+x\cdot 3(x-4{)}^2\cdot 1\\ & =(x-4{)}^3+3x(x-4{)}^2\\ & =(x-4{)}^2(x-4+3x)\\ f^{\prime }(x)& =(x-4{)}^2(4x-4)\end{array}$

Differentiate again with respect to $x$ using the product rule: $\begin{array}{rl}{f}^{\prime \prime }(x)& =2(x-4)\cdot 1\cdot (4x-4)+(x-4{)}^2\cdot 4\\ & =2(x-4)(4x-4)+4(x-4{)}^2\\ & =2\left(4x^2-4x-16x+16\right)+4\left(x^2-8x+16\right)\\ & =2\left(4x^2-20x+16\right)+4\left(x^2-8x+16\right)\\ & =8x^2-40x+32+4x^2-32x+64\\ & =12x^2-72x+96\\ f^{\prime \prime }(x)& =12\left(x^2-6x+8\right)\end{array}$

Set $f^{\prime \prime }(x)=0$: $\begin{array}{rl}12\left(x^2-6x+8\right)& =0\\ x^2-6x+8& =0\\ x^2-4x-2x+8& =0\\ x(x-4)-2(x-4)& =0\\ (x-4)(x-2)& =0\\ x& =4,x=2\end{array}$

The intervals of concavity are $(-\infty ,2)$, $(2,4)$, and $(4,\infty )$.

Case I: Consider the interval $(-\infty ,2)$ Let $x=1\in (-\infty ,2)$: $\begin{array}{rl}{f}^{\prime \prime }(1)& =12(1-6+8)\\ & =12(3)\\ & =36>0\end{array}$

Since $f^{\prime \prime }(x)>0$, the function is concave upward on $(-\infty ,2)$.

Case II: Consider the interval $(2,4)$ Let $x=3\in (2,4)$: $\begin{array}{rl}{f}^{\prime \prime }(3)& =12\left(9-18+8\right)\\ & =12(-1)\\ & =-12<0\end{array}$

Since $f^{\prime \prime }(x)<0$, the function is concave downward on $(2,4)$.

Case III: Consider the interval $(4,\infty )$ Let $x=5\in (4,\infty )$: $\begin{array}{rl}{f}^{\prime \prime }(5)& =12\left(25-30+8\right)\\ & =12(3)\\ & =36>0\end{array}$

Since $f^{\prime \prime }(x)>0$, the function is concave upward on $(4,\infty )$.

Result: Concave upward on $(-\infty ,2)\cup (4,\infty )$; concave downward on $(2,4)$.

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Part (v): $f(x)=x^{\frac{1}{2}}+2x$

Given: $f(x)=x^{\frac{1}{2}}+2x=\sqrt{x}+2x$

Note: Domain is $x\ge 0$, but for the second derivative we consider $x>0$.

Differentiate with respect to $x$: $f^{\prime }(x)=\frac{1}{2}{x}^{-\frac{1}{2}}+2$

Differentiate again with respect to $x$: $\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{1}{2}\left(-\frac{1}{2}{x}^{-\frac{3}{2}}\right)\\ f^{\prime \prime }(x)& =\frac{-1}{4x^{\frac{3}{2}}}\end{array}$

Since $x^{\frac{3}{2}}>0$ for all $x>0$, we have $f^{\prime \prime }(x)<0$ for all $x>0$.

Result: Concave downward on $(0,\infty )$; never concave upward on its domain.

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Part (vi): $f(x)=x+\frac{9}{x}$

Given: $f(x)=x+\frac{9}{x}$

Note: Domain is $x\ne 0$.

Differentiate with respect to $x$: $\begin{array}{rl}{f}^{\prime }(x)& =1+9\cdot \left(\frac{-1}{x^2}\right)\\ f^{\prime }(x)& =1-9x^{-2}\end{array}$

Differentiate again with respect to $x$: $\begin{array}{rl}{f}^{\prime \prime }(x)& =-9\left(-2x^{-3}\right)\\ f^{\prime \prime }(x)& =18x^{-3}\\ f^{\prime \prime }(x)& =\frac{18}{x^3}\end{array}$

Analyze the sign:

• For $x<0$: $x^3<0$, so $f^{\prime \prime }(x)=\frac{18}{x^3}<0$
• For $x>0$: $x^3>0$, so $f^{\prime \prime }(x)=\frac{18}{x^3}>0$

Result: Concave downward on $(-\infty ,0)$; concave upward on $(0,\infty )$.

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Key Formulas or Methods Used

• Second Derivative Test for Concavity:

  • If $f^{\prime \prime }(x)>0$ on an interval, $f$ is concave upward (convex) on that interval
  • If $f^{\prime \prime }(x)<0$ on an interval, $f$ is concave downward (concave) on that interval

• Differentiation Rules:

  • Power Rule: $\frac{d}{dx}[x^n]=n{x}^{n-1}$
  • Product Rule: $\frac{d}{dx}[u\cdot v]=u^{\prime }v+u{v}^{\prime }$
  • Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$

• Interval Testing Method: Choose test points in intervals determined by roots of $f^{\prime \prime }(x)=0$ or points where $f^{\prime \prime }(x)$ is undefined

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Summary of Steps

1. Find the second derivative $f^{\prime \prime }(x)$ by differentiating the function twice
2. Identify critical points where $f^{\prime \prime }(x)=0$ or $f^{\prime \prime }(x)$ is undefined (these are potential inflection points)
3. Divide the domain into intervals using the critical points found in step 2
4. Test each interval by selecting a sample point and evaluating the sign of $f^{\prime \prime }(x)$ at that point
5. Determine concavity:
   • $f^{\prime \prime }(x)>0\Rightarrow$ concave upward
   • $f^{\prime \prime }(x)<0\Rightarrow$ concave downward
6. State the final answer as intervals with the corresponding concavity direction