# How do you determine whether the function f(x)= 2x^3-3x^2-36x-7 is concave up or concave down and its intervals?

Aug 3, 2015

You can use the second derivative test.

#### Explanation:

The second derivative test allows you to determine the intervals on which a function is concave up or concave down by examining the sign of the second derivative around the inflexion point(s).

Inflexion point(s) are determined by making the second derivative equal to zero.

If the second derivative is positive on a given interval, then the function will be concave up on the same interval. Likewise, if the second derivative is negative on a given interval, the function will be concave down on said interval.

So, calculate the first derivative first - use the power rule

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 {x}^{2} - 36 x - 7\right)$

${f}^{'} = 6 {x}^{2} - 6 x - 36$

Next, calculate the second derivative. Once again, use the power rule

$\frac{d}{\mathrm{dx}} \left({f}^{'} \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(6 {x}^{2} - 6 x - 36\right)$

${f}^{' '} = 12 x - 6$

Find the inflexion point by solving ${f}^{' '} = 0$

$12 x - 6 = 0 \implies x = \textcolor{g r e e n}{\frac{1}{2}}$

Now look on how the second derivative behaves for values of $x$ smaller than $\frac{1}{2}$ and larger than $\frac{1}{2}$.

The two intervals you're going to use are

• $\left(- \infty , \frac{1}{2}\right)$

For this interval, ${f}^{' '}$ will be negative, which means that $f \left(x\right)$ is concave down.

• $\left(\frac{1}{2} , + \infty\right)$

For this interval, ${f}^{' '}$ will be positive, which implies that $f \left(x\right)$ is concave up.

So, $f \left(x\right)$ is concave down on $\left(- \infty , \frac{1}{2}\right)$ and concave up on $\left(\frac{1}{2} , + \infty\right)$. The point $\left(\frac{1}{2} , f \left(\frac{1}{2}\right)\right)$ is the only inflexion point on the graph of $f \left(x\right)$.