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Given the function 
h(x)=x^(2)-5x+2, determine the average rate of change of the function over the interval 
2 <= x <= 9.
Answer:

Given the function h(x)=x25x+2 h(x)=x^{2}-5 x+2 , determine the average rate of change of the function over the interval 2x9 2 \leq x \leq 9 .\newlineAnswer:

Full solution

Q. Given the function h(x)=x25x+2 h(x)=x^{2}-5 x+2 , determine the average rate of change of the function over the interval 2x9 2 \leq x \leq 9 .\newlineAnswer:
  1. Calculate Lower Endpoint: To find the average rate of change of the function h(x)=x25x+2h(x) = x^2 - 5x + 2 over the interval [2,9][2, 9], we need to calculate the difference in the function values at the endpoints of the interval and divide by the difference in the xx-values.
  2. Calculate Upper Endpoint: First, we calculate the value of h(x)h(x) at the lower endpoint of the interval, which is x=2x = 2.\newlineh(2)=(2)25(2)+2=410+2=4h(2) = (2)^2 - 5(2) + 2 = 4 - 10 + 2 = -4.
  3. Find Average Rate of Change: Next, we calculate the value of h(x)h(x) at the upper endpoint of the interval, which is x=9x = 9. h(9)=(9)25(9)+2=8145+2=38h(9) = (9)^2 - 5(9) + 2 = 81 - 45 + 2 = 38.
  4. Find Average Rate of Change: Next, we calculate the value of h(x)h(x) at the upper endpoint of the interval, which is x=9x = 9. h(9)=(9)25(9)+2=8145+2=38h(9) = (9)^2 - 5(9) + 2 = 81 - 45 + 2 = 38. Now, we find the average rate of change by subtracting the function value at the lower endpoint from the function value at the upper endpoint and dividing by the difference in xx-values. Average rate of change = (h(9)h(2))/(92)=(38(4))/(92)=42/7=6(h(9) - h(2)) / (9 - 2) = (38 - (-4)) / (9 - 2) = 42 / 7 = 6.

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