# What are the critical points of f(t) = e^tsqrt(2-t+t^2)?

Jun 11, 2018

The function $f \left(t\right)$ has no critical points.

#### Explanation:

A critical point on a function occurs whenever the derivative function, $f ' \left(t\right)$, is equal to $0$.

To find the critical points, we find the derivative of $f \left(t\right)$.
$f \left(t\right) = {e}^{t} \sqrt{{t}^{2} - t + 2}$

Use the product rule, which states that
$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)}$

$f ' \left(t\right) = \left({e}^{t}\right) \left(\frac{2 t - 1}{2 \sqrt{{t}^{2} - t + 2}}\right) + \left({e}^{t}\right) \left(\sqrt{{t}^{2} - t + 2}\right)$

To set $f ' \left(t\right)$ equal to zero, we factor $f ' \left(t\right)$:
$f ' \left(t\right) = \left({e}^{t}\right) \left(\frac{2 t - 1}{2 \sqrt{{t}^{2} - t + 2}} + \sqrt{{t}^{2} - t + 2}\right)$

Find a common denominator for the second fraction:
$= \left({e}^{t}\right) \left(\frac{2 t - 1}{2 \sqrt{{t}^{2} - t + 2}} + \frac{\sqrt{{t}^{2} - t + 2} \cdot \textcolor{b l u e}{2 \sqrt{{t}^{2} - t + 2}}}{\textcolor{b l u e}{2 \sqrt{{t}^{2} - t + 2}}}\right)$

$= \left({e}^{t}\right) \left(\frac{2 t - 1 + 2 \left({t}^{2} - t + 2\right)}{2 \sqrt{{t}^{2} - t + 2}}\right)$

$= \left({e}^{t}\right) \left(\frac{\cancel{2 t} - 1 + 2 {t}^{2} \cancel{- 2 t} + 4}{2 \sqrt{{t}^{2} - t + 2}}\right)$

$f ' \left(t\right) = \left({e}^{t}\right) \left(\frac{2 {t}^{2} + 3}{2 \sqrt{{t}^{2} - t + 2}}\right)$

Now, set this equal to zero. Use the zero product rule:
${e}^{t} \ne 0$

$0 \ne 2 {t}^{2} + 3$

Because $f ' \left(t\right)$ is never equal to zero, the function $f \left(t\right)$ has no critical points.