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Find the approximate change in T for the function T=4+3u-3u^2 when u is increased by 5% from the value of 2.
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Find \(\Large \frac{dT}{du}\) Replace \(\Large \frac{dT}{du}\) with \(\Large \frac{\Delta T}{\Delta u}\) Substitute, \(\Delta u = 0.05, u = 2\) and find the approximate value of \(\Delta T\).
can u pls tell us the exact answer for the problem @aum
Another approach: \[u\xrightarrow{\large\text{5% increase}}u+0.05u=1.05u\] So when \(u=2\), you have \(1.05u=2.1\). When \(u=2\), you have \[T=4+3(2)-3(2)^2=-2\] When \(u=2.1\), \[T=4+3(2.1)-3(2.1)^2=-2.93\] Find the percent change in the \(T\) values.
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