Question

graph the cube root function and analyze the minimum and maximum on the given interval. ( f(x) = sqrt3{x + 2}, -10,3 ) the minimum on the interval -10,3 is -2.00. (round to two decimal places as needed.) the maximum on the interval -10,3 is (square). (round to two decimal places as needed.) help me solve this view an example get more help

Explanation:

Step1: Analyze the function \( f(x)=\sqrt[3]{x + 2} \)

The cube - root function \( y = \sqrt[3]{u} \) is a monotonically increasing function for all real numbers \( u \), because the derivative of \( y=\sqrt[3]{u}=u^{\frac{1}{3}} \) is \( y^\prime=\frac{1}{3}u^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{u^{2}}} \), and for \( u
eq0 \), \( y^\prime>0 \), and at \( u = 0 \), the function is continuous and has a vertical tangent. In our function \( f(x)=\sqrt[3]{x + 2} \), let \( u=x + 2 \), and \( u=x + 2 \) is also a linear function with a slope of \( 1>0 \), so the composition \( f(x)=\sqrt[3]{x + 2} \) is a monotonically increasing function on the entire real - number line.

Step2: Find the maximum on the interval \([- 10,3]\)

Since \( f(x)\) is monotonically increasing on \([-10,3]\), the maximum value of the function on the interval \([a,b]\) (where \( a=-10 \) and \( b = 3 \)) occurs at the right - hand endpoint of the interval.
We substitute \( x = 3 \) into the function \( f(x)=\sqrt[3]{x + 2} \):
\( f(3)=\sqrt[3]{3 + 2}=\sqrt[3]{5}\approx1.71 \) (rounded to two decimal places)

Answer:

The maximum on the interval \([-10,3]\) is \( 1.71 \)