It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.

# Calculus

## Second Derivatives

In our previous sections we discussed the first derivative of a function and how to determine intervals of increase and decrease. The second derivative of a function, written as $$f''(x)$$ or $$\frac{d^2y}{d^2x}$$, can help us determine when the first derivative is increasing or decreasing and consequently the points of inflection in the graph of our original function.

If the second derivative is positive $$\Longrightarrow$$ the first derivative is increasing $$\Longrightarrow$$ the slope of the tangent line to the function is increasing as x increases. We see this phenomenon graphically as the curve of the graph being concave up, that is, shaped like a parabola open upward.

Likewise, if the second derivative is negative $$\Longrightarrow$$ the first derivative is decreasing $$\Longrightarrow$$ the slope of the tangent line to the function is decreasing as x increases. Graphically, we see this as the curve of the graph being concave down, that is, shaped like a parabola open downward. At the points where the second derivative is zero, we do not learn anything about the shape of the graph: it may be concave up or concave down, or it may be changing from concave up to concave down or changing from concave down to concave up.

To summarize:

$$\rightarrow f''(x) > 0$$ at $$x = p \Longrightarrow f(x)$$ is concave up at $$x = p$$

$$\rightarrow f''(x) < 0$$ at $$x = p \Longrightarrow f(x)$$  is concave down at $$x = p$$

$$\rightarrow f''(x) = 0$$ at $$x = p \Longrightarrow$$ we do not know anything new about the behaviour of $$f(x)$$ at $$x = p$$

Consider $$f(x) = 3x^3 - 6x^2 + 2x -1$$. Then $$f'(x) = 9x^2 -12x + 2$$ and $$f''(x) = 18x - 12$$. Let us look at the values $$x = 0$$ and $$x = 1$$:

$$\Longrightarrow f''(0) = -12 < 0$$ and thus we know that the graph of $$f(x)$$ is concave down at this point.

$$\Longrightarrow f''(1) = 6 > 0$$ and thus we know that the graph of $$f(x)$$ is concave up at this point.

## Points of Inflection

A function $$f(x) \( has an inflection point at \( x = p$$ if the graph of the function changes concavity (concave up to concave down, or vice versa) at that point. It is clear, then, that an inflection point can only happen when the second derivative is 0, because otherwise the point would the graph would be either completely concave up or completely concave down as we learned in the section above. Just like in the case of local maxima and local minima and the first derivative, however, the presence of a point where the second derivative of a function is 0 does not automatically tell us that the point is an inflection point.

Consider $$f(x) = x^4$$. Then $$f’(x) = 4x^3$$ and $$f''x)=12x^2$$.

$$f''(0) = 0$$, but if we were sketch the function $$f(x) = x^4$$, it becomes clear that $$x = 0$$ is not an inflection point: So while the second derivative can tell us a lot about the shape of the graph of a function, it cannot tell us if the graph of a function has an inflection point; it can only tell us where it might have an inflection point.

EXAMPLE

Determine the intervals of concavity for the graph of the function $$f(x) = \frac{3}{20}x^5 - \frac{2}{3}x^4 + \frac{7}{6}x^3 - x^2$$.

See the video below for the solution: