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

Question

Luke Alvarez · Accepted Answer

The function #f(t)# has no critical points.

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.

Axel Alvarez · Answer

undefined To find the critical points of $ f(t) = e^{t}\sqrt{2 - t + t^2} $, we need to find where the derivative $ f'(t) $ is zero or undefined.

First, calculate the derivative $ f'(t) $:

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

Set $ f'(t) $ to zero and solve for $ t $:

$$ e^{t}\sqrt{2 - t + t^2} + e^{t}\frac{1}{2\sqrt{2 - t + t^2}}(-1 + 2t) = 0 $$

After solving this equation, you will find the critical points of $ f(t) $.