Question

question 22 (1 point) what is the minimum value for the function f(x)=4.5x^2 - 7.1x + 3.9? round to two decimal places. a) 6.70 b) 1.10 c) 3.90 d) 0.79

Explanation:

Step1: Find the derivative

For the function $y = 4.5x^{2}-7.1x + 3$, the derivative $y^\prime=9x - 7.1$

Step2: Set the derivative equal to 0

Set $y^\prime = 0$, so $9x-7.1 = 0$. Solving for $x$ gives $x=\frac{7.1}{9}\approx0.79$

Step3: Substitute $x$ into the original function

Substitute $x = \frac{7.1}{9}$ into $y = 4.5x^{2}-7.1x + 3$.
\[

$$\begin{align*} y&=4.5\times(\frac{7.1}{9})^{2}-7.1\times\frac{7.1}{9}+3\\ &=4.5\times\frac{50.41}{81}-\frac{50.41}{9}+3\\ &=\frac{226.845}{81}-\frac{453.69}{81}+3\\ &=\frac{226.845 - 453.69+243}{81}\\ &=\frac{226.845+243 - 453.69}{81}\\ &=\frac{469.845 - 453.69}{81}\\ &=\frac{16.155}{81}\\ &\approx0.79 \end{align*}$$

\]

Answer:

d) 0.79