Concavity

The 1st derivative is

$$f^{\prime }(x)=x^4+x^3$$

To find the critical points, set equal to $0$:

$$\begin{gathered} x^4+x^3=0 \\ {x}^3(x+1)=0 \\ x=0,-1 \end{gathered}$$

Now apply the 2nd derivative test. The 2nd derivative is

$$f^{\prime \prime }(x)=4x^3+3x^2$$

Evaluate each critical point:

$$\begin{array}{rl}{f}^{\prime \prime }(-1)& =-1\\ f^{\prime \prime }(0)& =0\end{array}$$

Classify each:

1. Since $f^{\prime \prime }(-1)<0$, $f(x)$ is concave down at $x=-1$, so it is a relative max.

2. Since $f^{\prime \prime }(0)=0$, the test is inconclusive. Using the 1st derivative test,

Interval	Sign of $f^{\prime }$
$x<0$	$-$
$x>0$	$+$

Since $f^{\prime }$ changes from negative to positive, there is a relative min at $x=0$.

Critical points of $f$ occur where $f^{\prime }(x)=0$ or is undefined.

As a calculator-active question, you can simply define the given derivative of $f$ as a separate function $g$ in Desmos by typing the following:

$$g(x)=2x^2-e^{\cos (x)}$$

Since $g(x)=f^{\prime }(x)$, note where the graph of $g$ crosses the $x$-axis, which is where $f^{\prime }(x)=0$. These are the critical points of $x=\pm 0.947$.

Because $g(x)=f^{\prime }(x)$, then differentiating $g(x)$ gives the 2nd derivative of $f$ i.e. $g^{\prime }(x)=f^{\prime \prime }(x)$.

Instead of differentiating by hand, you can use Desmos to evaluate the 2nd derivative at your critical points automatically. Simply type these two lines:

$$g^{\prime }(-0.947)$$

$$g^{\prime }(0.947)$$

which results in:

$$\begin{array}{rl}{g}^{\prime }(-0.947)& =-5.243\\ g^{\prime }(0.947)& =5.243\end{array}$$

Analysis:

• Since $f^{\prime \prime }(-0.947)<0$, then $x=-0.947$ is a relative maximum of $f$

• Since $f^{\prime \prime }(0.947)>0$, then $x=0.947$ is a relative minimum of $f$.

Note that to use the 1st derivative test to classify extrema on $f$ given the derivative $f^{\prime }$, simply observe the graph of $g$ at the critical points. For example, at $x=-0.947$, the graph of $g=f^{\prime }$ changes from positive to negative, confirming that $f$ has a relative maximum at that critical point.

To classify the extrema at $x=1$, we need to apply the 2nd derivative test and determine the sign of $G^{\prime \prime }(1)$. Differentiating $G$ with the chain rule,

$$\begin{array}{r}{G}^{\prime }(x)=-2f(2x)\cdot f^{\prime }(2x)\cdot 2\\ =-4f(2x)\cdot f^{\prime }(2x)\end{array}$$

We can confirm that $x=1$ is a critical point since $G^{\prime }(1)=-4f(2)f^{\prime }(2)=0$ (given that $f^{\prime }(2)=0$).

Next, differentiate $G^{\prime }$. using the product and chain rules:

$$G^{\prime \prime }(x)=-4\left(f(2x)\cdot 2f^{\prime \prime }(2x)+f^{\prime }(2x)\cdot 2f^{\prime }(2x)\right)$$

1. Find potential points of tangency:

The curve meets $y=-2$, which is a horizontal tangent line with slope $0$. Therefore, at any tangency points $(a,-2)$, we must have

$$\frac{dy}{dx}=0.$$

The first derivative

$$\frac{dy}{dx}=\frac{x^2-4x+3}{y+5}$$

is $0$ when the numerator equals $0$:

$$\begin{gathered} x^2-4x+3=0 \\ (x-1)(x-3)=0 \\ x=1,3 \end{gathered}$$

$(1,-2)$ and $(3,-2)$ are both points where the tangent line could be horizontal.

2. Find the 2nd derivative:

Differentiating $\frac{dy}{dx}$ implicitly,

$$\frac{d^{2}y}{d{x}^2}=\frac{(y+5)(2x-4)-(x^2-4x+3)(\frac{dy}{dx})}{(y+5{)}^2}$$

3. Classify the points with the 2nd derivative test:

At any point of horizontal tangency, $\frac{dy}{dx}=0$ and $x^2-4x+3=0$. This eliminates the entire trailing term in the numerator:

$$\frac{d^{2}y}{d{x}^2}=\frac{(y+5)(2x-4)-0}{(y+5{)}^2}=\frac{2x-4}{y+5}$$

Substitute each candidate point into this simplified expression:

• For $(1,-2)$:

$$\frac{d^{2}y}{d{x}^2}=\frac{2(1)-4}{-2+5}=\frac{-2}{3}$$

Since $\frac{d^{2}y}{d{x}^2}<0$, the curve is concave down and $(1,-2)$ is a relative maximum.

• For $(3,-2)$:

$$\frac{d^{2}y}{d{x}^2}=\frac{2(3)-4}{-2+5}=\frac{2}{3}$$

Since $\frac{d^{2}y}{d{x}^2}>0$, the curve is concave up and $(3,-2)$ is a relative minimum.