Find all critical point(s) of the function 𝑓(𝑡)= t-6sqrt (t+1)

Answer 1

f(𝑡) = t - 6 √( t + 1 )

[ √( t + 1 ) ] ′ = d [ √( t + 1 ) ] / dt

Using the chain rule:

d [ √( t + 1 ) ] / dt = d√u / du ∙ du / dt

where u = t + 1

d [ √( t + 1 ) ] / dt = d (u )¹′² / du ∙ d ( t +1 ) / dt = 1 / 2 t ⁻ ¹′² ∙ 1 = 1 / 2√( t + 1 )

f(𝑡)′ = t ′ - 6 ∙ d [ √( t + 1 ) ] / dt = 1 - 6 ∙ 1 / 2√( t + 1 ) = 1 - 3 /√( t + 1 )

The point c is called a critical point if f ′(c) = 0 or f ′(c) does not exist.

In this case:

f(𝑡)′ = 0

1 - 3 /√( t + 1 ) = 0

Add 3 /√( t + 1 ) to both sides

1 = 3 /√( t + 1 )

Multiply both sides by

√( t + 1 )

√( t + 1 ) = 3

Raise both sides by power of two

t + 1 = 9

Subtract 1 to both sides

t = 8

f ′(c) does not exist when t = - 1 because:

1 - 3 /√( t + 1 ) = 1 - 3 /√( -1 + 1 ) = 1 - 3 /√( 0 ) = 1 - 3 / 0 = undefined

So critical points of the f(𝑡) = t - 6 √( t + 1 ) are:

t = - 1 and t = 8