When x = t^{2} + t and y = e^{t} + 1, for what values of t is the curve concave upwards?

Question:

When {eq}\,x = t^{2} + t \; {/eq} and {eq}\; y = e^{t} + 1\, {/eq}, for what values of {eq}t\, {/eq} is the curve concave upwards?

Concavity and Inflection Points

A function is concaving up when its double derivative is positive i.e. {eq}\dfrac{d^2y}{dx^2}>0 {/eq}, and it is concaving down when {eq}\dfrac{d^2y}{dx^2}<0 {/eq}

The point where the concavity of the function changes is called the inflection point of the function.

{eq}x = t^{2} + t\\ y = e^{t} + 1\\ \dfrac{dy}{dx} = \bigg(\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\bigg )\\ \dfrac{dy}{dx} = \dfrac {e^{t}}{2t+1}\\ \dfrac{d^2y}{dx^2} = \dfrac {d(\frac{dy}{dx})}{dx}\\ \dfrac{d^2y}{dx^2} = \dfrac {d(\frac{dy}{dx})}{dt}\times \dfrac{dt}{dx}\\ \dfrac{d^2y}{dx^2} = \dfrac{(2t+1)e^t - 2e^t}{(2t+1)^2} \times \dfrac 1{2t+1}\\ \dfrac{d^2y}{dx^2} = \dfrac{(2t-1)e^t}{(2t+1)^3} {/eq}

For concaving up, {eq}\dfrac{d^2y}{dx^2}>0 {/eq}

{eq}\Rightarrow \dfrac{(2t-1)e^t}{(2t+1)^3} > 0 {/eq}

It is positive when numerator and denominator are either both positive or both negative.

Numerator is positive for t > 0.5 and negative for t < 0.5

Denominator is positive for t > -0.5 and negative for t < -0.5

So function is concaving up for t > 0.5 and t < -0.5