Question

f(x) = \sqrt3{x} + 2, -10,3 the minimum on the interval -10,3 is \boxed{}. (round to two decimal places as needed.)

Explanation:

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

The function \( y=\sqrt[3]{u} \) is a cube - root function, and for \( u=x + 2 \), the cube - root function is increasing for all real numbers \( u \) (since the derivative of \( y=\sqrt[3]{u} \) with respect to \( u \) is \( y^\prime=\frac{1}{3u^{\frac{2}{3}}}\geq0 \) for all \( u
eq0 \), and at \( u = 0 \), the function is continuous). The function \( u=x + 2 \) is also a linear function with a slope of 1 (increasing function). By the composition of functions, if \( y = f(u) \) is increasing and \( u = g(x) \) is increasing, then \( y=f(g(x)) \) is increasing. So \( f(x)=\sqrt[3]{x + 2} \) is an increasing function on the interval \([-10,3]\).

Step2: Find the minimum value

For an increasing function on an interval \([a,b]\), the minimum value occurs at the left - hand endpoint of the interval. Here, the interval is \([-10,3]\), so we evaluate the function at \( x=-10 \).

Substitute \( x=-10 \) into the function \( f(x)=\sqrt[3]{x + 2} \):

\( f(-10)=\sqrt[3]{-10 + 2}=\sqrt[3]{-8}=- 2.00 \) (rounded to two decimal places)

Answer:

\(-2.00\)