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

1 Answer
Jun 11, 2018

Answer:

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.