Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Given the function h(x) = x² − 10x + 21, determine the average rate of change of the function over the interval -1 ≤ x ≤ 10.

Sagot :

Answer:  the average rate of change of the function is 23.

Step-by-step explanation:

[tex]h(x)=x^2-10x+21\ \ \ \ \ -1\leq x\leq 10 \ \ \ \ A(x)=?\\\displaystyle\\A(x)=\frac{\Delta y}{\Delta x}=\frac{h(x-g)-h(x)}{g} \\g=x_{max}-x_{min}\\g=10-(-1)\\g=10+1\\g=11.\ \ \ \ \ \Rightarrow\\A(x)=\frac{(x-11)^2-10*(x-11)+21-(x^2-10x+21)}{11} \\A(x)=\frac{x^2-22x+121-10x+110+21-x^2+10x-21}{11} \\A(x)=\frac{231-22x}{11}\\ A(x)=\frac{11*(21-2x)}{11}\\ A(x)=21-2x.\\x=x_{min}\ \ \ \ \ \Rightarrow\\A(x)=21-2*(-1)\\A(x)=21+2\\A(x)=23.[/tex]

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.