Final Answer:
The estimated value of
for
is ( 15.000 ) (rounded to three decimal places).
Step-by-step explanation:
To estimate
, we can use the definition of the derivative:
![\[ g'(t) = \lim_{{h \to 0}} (g(t + h) - g(t))/(h) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/4pxu7uhmv1rld874vanuhbqn0jtbvewn4s.png)
Apply this formula to
:
![\[ g'(t) = \lim_{{h \to 0}} ((3)/((t + h)^5) - (3)/(t^5))/(h) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/8n02v3pyh4zgn5e9md7ox6s5isf7bjlt5v.png)
Now, substitute ( t = 1 ) into this expression:
![\[ g'(1) = \lim_{{h \to 0}} ((3)/((1 + h)^5) - (3)/(1^5))/(h) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/n4vfrkglgz5wnvq302sphmjd2bxuos6ycv.png)
Simplify further:
\
![[ g'(1) = \lim_{{h \to 0}} ((3)/((1 + h)^5) - 3)/(h) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/vvh9s17ceaa2yfbtnchzyz75n5wgcua5gx.png)
Now, compute the limit. As ( h ) approaches 0, the expression becomes
, which is an indeterminate form. Apply L'Hôpital's Rule, taking the derivative of the numerator and denominator with respect to ( h ):
![\[ g'(1) = \lim_{{h \to 0}} ((d)/(dh)\left((3)/((1 + h)^5) - 3\right))/((d)/(dh)h) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/b6vis8i5lv4nl7c6bkt2v0byjd5jeasnv9.png)
After evaluating this expression, we find (g'(1) = 15.000) (rounded to three decimal places).