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

1 Answer
Jun 11, 2018

The function f(t) has no critical points.

Explanation:

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

To find the critical points, we find the derivative of f(t).
f(t) = e^t sqrt(t^2-t+2)

Use the product rule, which states that
color(blue)(d/dx (f(x)*g(x)) = f(x)g'(x) + f'(x)g(x))

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

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

Find a common denominator for the second fraction:
= (e^t)(frac{2t-1}{2sqrt(t^2-t+2)}+frac{sqrt(t^2-t+2)*color(blue)(2sqrt(t^2-t+2))}{color(blue)(2sqrt(t^2-t+2))})

= (e^t)(frac{2t-1 + 2(t^2-t+2)}{2sqrt(t^2-t+2)})

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

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

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

0 ne 2t^2+3

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