# Find the critical values of the function f(t)=t*square root(1-t), if t<1?

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The critical values of a function f(x) can be derived by solving f'(x) = 0 for x.

f(t)=t*sqrt(1-t)

f'(t) = t'*sqrt(1 - t) + t*[sqrt (1- t)]'

=> f'(t) = sqrt (1 - t) + t*(1/2)(-1)/sqrt (1-t)

sqrt (1 - t) + t*(1/2)(-1)/sqrt (1-t) = 0

=> 1 - tÂ - t/2 = 0

=> 1-3t/2 = 0

=> t = 2/3

**The critical value of f(t) lies at t = 2/3**

The critical values of a function are the roots of the 1st derivative of the function.

We'll differentiate the function with respect to t, using the product rule:

f'(t) = sqrt(1-t) - t/2sqrt(1-t)

We'll cancel f'(t):

f'(t) = 0

sqrt(1-t) - t/2sqrt(1-t) = 0

2(1-t) - t = 0

We'll remove the brackets:

2 - 2t - t = 0

We'll combine like terms:

-3t = -2

t = 2/3

**The critical value of the given function is t = 2/3.**