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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]

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