How to solve the equation uov(t)=0? u(t)=2t^2-1 v(t)=t-1

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have the functions u(t) = 2t^2 - 1 and v(t) = t -1.

uov(t) = u(v(t))

v(t) = t - 1

substituting t - 1 in u(t)

=> 2*( t - 1)^2 - 1

As uov(t) = 0

=> 2*( t - 1)^2 - 1 = 0

=> 2( t^2 + 1 - 2t) - 1 = 0

=> 2t^2 + 2 - 4t - 1=0

=> 2t^2 - 4t + 1 =0

t1 = [ 4 + sqrt ( 16 - 8)]/4

=> t1 = 1 + sqrt 2 /2 = 1.707

t2 = [ 4 - sqrt ( 16 - 8)]/4

=> t1 = 1 - sqrt 2 /2 = .2928

Therefore t can take on the values 1 + [(sqrt 2) / 2] and 1 - [(sqrt 2) /2].

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To solve the equation means to determine t, we'll have to find (uov)(t).

(uov)(t) = u(v(t))

We'll substitute t by v(t):

u(v(t)) = 2[v(t)]^2 - 1

u(v(t)) = 2(t-1)^2 - 1

We'll expand the square:

u(v(t)) = 2t^2 - 4t + 2 - 1

We'll combine like terms:

u(v(t)) = 2t^2 - 4t + 1

We'll put u(v(t)) = 0:

2t^2 - 4t + 1 = 0

We'll apply quadratic formula:

t1 = [4+sqrt(16 - 8)]/4

t1 = (4+2sqrt2)/4

t1 = (2+sqrt2)/2

t2 = (2-sqrt2)/2

The values of t for u(v(t)) = 0 are: {(2-sqrt2)/2 ; (2+sqrt2)/2}

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