Find the interval where the function g(x) = \int_{0}^{x} \frac{dt}{t^{2} - t + 1} is concave up.

Question:

Find the interval where the function

{eq}\displaystyle\; g(x) = \int_{0}^{x} \frac{dt}{t^{2} - t + 1} {/eq}

is concave up.

Condition for checking Concave Upwards and downwards :

Let us consider any one variable function: {eq}y = f(x) {/eq} then,

{eq}f(x) {/eq} is concave upwards if {eq}f''(x) > 0 {/eq}

Again: {eq}f(x) {/eq} is concave downwards if {eq}f''(x) < 0 {/eq}

Answer and Explanation:

Here, in this case, the given function is {eq}\displaystyle\; g(x) = \int_{0}^{x} \frac{dt}{t^{2} - t + 1} {/eq}

By applying derivative with respect to {eq}x {/eq}, we get:

{eq}\eqalign{ \:g'(x)& = \frac{d}{{dx}}\int\limits_0^x {\frac{{dt}}{{{t^2} - t + 1}}} \cr & = \frac{1}{{{x^2} - x + 1}} \cr g''\left( x \right)& = \frac{d}{{dx}}\left( {\frac{1}{{{x^2} - x + 1}}} \right) \cr & = - \frac{{2x - 1}}{{{{\left( {{x^2} - x + 1} \right)}^2}}} \cr} {/eq}

Hence the given function is concave upwards, when we have:

{eq}\eqalign{ g''\left( x \right) > 0 & \Rightarrow - \frac{{2x - 1}}{{{{\left( {{x^2} - x + 1} \right)}^2}}} > 0 \cr & \Rightarrow - \left( {2x - 1} \right) > 0 \cr & \Rightarrow 1 - 2x > 0 \cr & \Rightarrow 2x < 1 \cr & \Rightarrow x < \frac{1}{2} \cr} {/eq}

Hence the required interval can be given as:

{eq}\displaystyle -\infty <x <\frac{1}{2} {/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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