Answer :
To find the critical numbers of the function [tex]\( g(t) = t \sqrt{7 - t} \)[/tex] for [tex]\( t < 6 \)[/tex], follow these steps:
1. Write the function:
[tex]\[ g(t) = t \sqrt{7 - t} \][/tex]
2. Find the derivative of the function:
To determine the critical points, we first need to find the derivative [tex]\( g'(t) \)[/tex].
3. Rewrite [tex]\( g(t) \)[/tex]:
[tex]\[ g(t) = t (7 - t)^{\frac{1}{2}} \][/tex]
Now, we use the product rule to differentiate [tex]\( g(t) \)[/tex]. The product rule states that if [tex]\( u(t) = t \)[/tex] and [tex]\( v(t) = (7 - t)^{\frac{1}{2}} \)[/tex], then:
[tex]\[ g'(t) = u'(t)v(t) + u(t)v'(t) \][/tex]
4. Differentiate [tex]\( u(t) \)[/tex] and [tex]\( v(t) \)[/tex]:
[tex]\[ u(t) = t \implies u'(t) = 1 \][/tex]
[tex]\[ v(t) = (7 - t)^{\frac{1}{2}} \implies v'(t) = \frac{1}{2} (7 - t)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{2} (7 - t)^{-\frac{1}{2}} \][/tex]
5. Apply the product rule:
[tex]\[ g'(t) = 1 \cdot (7 - t)^{\frac{1}{2}} + t \left(-\frac{1}{2} (7 - t)^{-\frac{1}{2}}\right) \][/tex]
Simplify this expression:
[tex]\[ g'(t) = (7 - t)^{\frac{1}{2}} - \frac{t}{2} (7 - t)^{-\frac{1}{2}} \][/tex]
[tex]\[ g'(t) = \frac{2(7 - t) - t}{2 (7 - t)^{\frac{1}{2}}} \][/tex]
[tex]\[ g'(t) = \frac{14 - 3t}{2 \sqrt{7 - t}} \][/tex]
6. Set the derivative equal to zero to find critical points:
[tex]\[ \frac{14 - 3t}{2 \sqrt{7 - t}} = 0 \][/tex]
This implies:
[tex]\[ 14 - 3t = 0 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 14 = 3t \][/tex]
[tex]\[ t = \frac{14}{3} \][/tex]
Therefore, the critical number of the function [tex]\( g(t) = t \sqrt{7 - t} \)[/tex], for [tex]\( t < 6 \)[/tex], is:
[tex]\[ t = \frac{14}{3} \][/tex]
So, the critical number is [tex]\( \frac{14}{3} \)[/tex].
1. Write the function:
[tex]\[ g(t) = t \sqrt{7 - t} \][/tex]
2. Find the derivative of the function:
To determine the critical points, we first need to find the derivative [tex]\( g'(t) \)[/tex].
3. Rewrite [tex]\( g(t) \)[/tex]:
[tex]\[ g(t) = t (7 - t)^{\frac{1}{2}} \][/tex]
Now, we use the product rule to differentiate [tex]\( g(t) \)[/tex]. The product rule states that if [tex]\( u(t) = t \)[/tex] and [tex]\( v(t) = (7 - t)^{\frac{1}{2}} \)[/tex], then:
[tex]\[ g'(t) = u'(t)v(t) + u(t)v'(t) \][/tex]
4. Differentiate [tex]\( u(t) \)[/tex] and [tex]\( v(t) \)[/tex]:
[tex]\[ u(t) = t \implies u'(t) = 1 \][/tex]
[tex]\[ v(t) = (7 - t)^{\frac{1}{2}} \implies v'(t) = \frac{1}{2} (7 - t)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{2} (7 - t)^{-\frac{1}{2}} \][/tex]
5. Apply the product rule:
[tex]\[ g'(t) = 1 \cdot (7 - t)^{\frac{1}{2}} + t \left(-\frac{1}{2} (7 - t)^{-\frac{1}{2}}\right) \][/tex]
Simplify this expression:
[tex]\[ g'(t) = (7 - t)^{\frac{1}{2}} - \frac{t}{2} (7 - t)^{-\frac{1}{2}} \][/tex]
[tex]\[ g'(t) = \frac{2(7 - t) - t}{2 (7 - t)^{\frac{1}{2}}} \][/tex]
[tex]\[ g'(t) = \frac{14 - 3t}{2 \sqrt{7 - t}} \][/tex]
6. Set the derivative equal to zero to find critical points:
[tex]\[ \frac{14 - 3t}{2 \sqrt{7 - t}} = 0 \][/tex]
This implies:
[tex]\[ 14 - 3t = 0 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 14 = 3t \][/tex]
[tex]\[ t = \frac{14}{3} \][/tex]
Therefore, the critical number of the function [tex]\( g(t) = t \sqrt{7 - t} \)[/tex], for [tex]\( t < 6 \)[/tex], is:
[tex]\[ t = \frac{14}{3} \][/tex]
So, the critical number is [tex]\( \frac{14}{3} \)[/tex].
