# Find the open intervals where the function f(x)=\ln(x^2+49) is concave upward or concave...

## Question:

Find the open intervals where the function {eq}f(x)=\ln(x^2+49) {/eq} is concave upward or concave downward. Find any inflection points.

## Concavity of a Function:

The concavity of a single variable function can be explained using the second derivative.

We find the point at which the second derivative is zero. Then we divide the entire real line using the zero of the second derivative.

After that, we check the sign of the second derivative in each of the intervals.

If the second derivative is positive on an interval, then the function is concave up on that interval. If the second derivative is negative on an interval, then the function is concave down on that interval.

To know about the concavity of a function, we need to equate the second derivative to zero.

We have {eq}f(x)=\ln(x^2+49) . {/eq}

Now:

{eq}f'(x)=\displaystyle \frac{1}{x^2+49} \ \frac{d}{dx}\left ( x^2+49 \right )=\frac{2x}{x^2+49}. {/eq}

{eq}\begin{align*} f''(x) &=\frac{d}{dx}\left ( \frac{2x}{x^2+49} \right )\\ \\ &=\frac{\left ( x^2+49 \right )\frac{d}{dx}\left ( 2x \right )-2x\frac{d}{dx}\left ( x^2+49 \right )}{\left ( x^2+49 \right )^2}\\ \\ &=\frac{2\left ( x^2+49 \right )-2x*2x}{\left ( x^2+49 \right )^2}\\ \\ &=\frac{ 2x^2+98-4 x^2}{\left ( x^2+49 \right )^2}\\ \\ f''(x)&=\frac{98-2x^2}{\left ( x^2+49 \right )^2}. \end{align*} {/eq}

{eq}f''(x)=0 \Rightarrow \displaystyle \frac{98-2x^2}{\left ( x^2+49 \right )^2}=0. {/eq}

{eq}\Rightarrow 98-2x^2=0 \Rightarrow x^2=49. {/eq}

That is {eq}x=\pm 7. {/eq}

Now consider the intervals {eq}\left ( -\infty ,-7 \right ), \ \left ( -7,7 \right ) {/eq} and {eq}\left ( 7,\infty \right ). {/eq}

Take any point from each of these intervals and check the sign of the second derivative.

Consider {eq}-8 \in \left ( -\infty ,-7 \right ), \ \displaystyle f''(-8)=\frac{98-2(-8)^2}{\left ( (-8)^2+49 \right )^2}<0. {/eq}

{eq}0 \in \left ( -7,7 \right ), \ \displaystyle f''(0)=\frac{98-2(0)^2}{\left ( (0)^2+49 \right )^2}>0. {/eq}

{eq}8 \in \left ( 7,\infty \right ), \ \displaystyle f''(8)=\frac{98-2(8)^2}{\left ( 8^2+49 \right )^2}<0. {/eq}

In general, {eq}f''(x)<0 {/eq} on {eq}\left ( -\infty ,-7 \right ) {/eq} and {eq}\left ( 7,\infty \right ) {/eq} and {eq}f''(x)>0 {/eq} on {eq}\left ( -7,7 \right ). {/eq}

Thus we can conclude that {eq}f(x) {/eq} is cocave up on {eq}\left ( -7,7 \right ) {/eq} and concave down on {eq}\left ( -\infty ,-7 \right ) \ {\&} \ \left ( 7,\infty \right ). {/eq}