Find the interval(s) on which the graph of y=f(x) is concave up if f(x) =\int_0^x...


Find the interval(s) on which the graph of {eq}y=f(x){/eq} is concave up if {eq}f(x) =\int_0^x \frac{1+t}{1+t^2}dt{/eq}.

Leibniz integral rule.

The most general form of differentiation under the integral sign states that: if f(x,t) is a continuous and continuously differentiable (i.e., partial derivatives exist and are themselves continuous) function and the limits of integration a(x) and b(x) are continuous and continuously differentiable functions of x, then

$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt= f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)dt $$

In the case where a(x) and b(x) are constant functions, this formula reduces to the simpler form

$$\frac{d}{dx}\int_{a}^{b} f(x,t)dt=\int_{a}^{b}\frac{\partial}{\partial x}f(x,t)dt $$

The graph of y = f (x) is concave upward on those intervals where y = f"(x) > 0. The graph of y = f (x) is concave downward on those intervals where y = f"(x) < 0.

Answer and Explanation:

{eq}\text{To find the intervals where the function f(x) is concave up we have to find the first and second derivative of f(x)}\\ {/eq}

$$\text{To find the first derivative of f(x) we will use Leibniz integral rule i.e. }\\ \frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt= f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)dt\\ f'(x) =\frac{d}{dx}\int_0^x \frac{1+t}{1+t^2}dt=\frac{1+x}{1+x^2}\cdot \frac{d}{dx}(x)-\frac{1+x}{1+x^2}\cdot \frac{d}{dx}(1)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\frac{1+t}{1+t^2}dt\\ f'(x)=\frac{1+x}{1+x^2}-0+0\\ f'(x)=\frac{1+x}{1+x^2}\\ \text{Now, calculate second derivative of f(x)}\\ f"(x)=\frac{(1+x^2)(1)-(1+x)(2x)}{(1+x^2)^2}\\ f"(x)=\frac{-x^2-2x+1}{(1+x^2)^2}\\ \text{equate }f"(x)=0\\ f"(x)=\frac{-x^2-2x+1}{(1+x^2)^2}=0\\ -x^2-2x+1=0\\ x^2+2x-1=0\\ x=-1\pm\sqrt2\\ \text{At }x<-1-\sqrt2,\;f"(x)<0,\\ \text{At } -1-\sqrt2>x<-1+\sqrt2,\;f"(x)>0,\\ \text{And at }x>-1+\sqrt2,\;f"(x)<0\\ \text{So, the function is concave down on the interval }(-\infty,-1-\sqrt2)U(-1+\sqrt2,\infty) \text{ and concave up on interval }(-1-\sqrt2,-1+\sqrt2)\\ $$

Learn more about this topic:

Understanding Concavity and Inflection Points with Differentiation

from Math 104: Calculus

Chapter 10 / Lesson 6

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