A function and its two derivatives are given. Find the intervals of concave up and concave down....

Question:

A function and its two derivatives are given. Find the intervals of concave up and concave down.

{eq}f(x)= \frac{5 - 4x}{4(x - 1)^2}, f'(x)= \frac{2x - 3}{2(x - 1)^3}, f"(x)= \frac{7 - 4x}{2(x - 1)^4} {/eq}

Concavity of a Function


If the graph of a function is given, we can determine the function's concavity, by looking where the tangent line to the graph lie with respect to the graph.

If the tangent line to the graph is above the graph, the function is concave down, otherwise, is concave up.

The concavity of a function, when the graph is not given, is determined by the sign of the second derivative:

{eq}\displaystyle \text{ if } f''(x)>0 \implies f(x) \text{ is concave up}\\ \displaystyle \text{ if } f''(x)<0 \implies f(x) \text{ is concave down}. {/eq}

We can simplify the calculations when solving inequalities that involve terms with powers, to eliminate the even powers terms, because they are positive all the time.

Answer and Explanation:


The function {eq}\displaystyle f(x)= \frac{5 - 4x}{4(x - 1)^2} {/eq} whose first and second derivatives are given as {eq}\displaystyle f'(x)= \frac{2x - 3}{2(x - 1)^3}, f"(x)= \frac{7 - 4x}{2(x - 1)^4} {/eq}

is concave up whenever the second derivative is positive.

Because the second derivative {eq}\displaystyle f"(x)= \frac{7 - 4x}{2(x - 1)^4} {/eq}

has the denominator {eq}\displaystyle 2(x - 1)^4 {/eq} positive for any value of {eq}\displaystyle x, {/eq} for being an even exponent power ,

the second derivative is positive whenever the numerator {eq}\displaystyle 7-4x {/eq} is positive,

therefore {eq}\displaystyle 7-4x>0\iff x<\frac{7}{4}\implies \boxed{\text{ the function is concave up on } \left(-\infty, \frac{7}{4}\right)} {/eq}

And the function is concave down whenever the second derivative is negative, so when the numerator is negative

{eq}\displaystyle 7-4x<0\iff x>\frac{7}{4}\implies \boxed{\text{ the function is concave down on } \left( \frac{7}{4},\infty\right)}. {/eq}


Learn more about this topic:

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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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