How do you find the average rate of change of #f(x) = 5x^2+2x-7# over the interval [5,7]?

Answer 1

#62#

#"the average rate of change of " f(x) " is"#
#(f(7)-f(5))/(7-5)larr(f(b)-f(a))/(b-a)to[a,b]=[5,7]#
#f(7)=5(7)^2+2(7)-7=252#
#f(5)=5(5)^2+2(5)-7=128#
#rArr"average rate of change "=(252-128)/(7-5)=62#
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Answer 2

To find the average rate of change of ( f(x) = 5x^2 + 2x - 7 ) over the interval ([5, 7]), you need to calculate the difference in the function values at the endpoints of the interval and divide by the difference in the x-values.

Here's the formula:

[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} ]

Where ( a ) and ( b ) are the endpoints of the interval.

In this case:

( f(5) = 5(5)^2 + 2(5) - 7 = 5(25) + 10 - 7 = 125 + 10 - 7 = 128 )

( f(7) = 5(7)^2 + 2(7) - 7 = 5(49) + 14 - 7 = 245 + 14 - 7 = 252 )

So, the average rate of change of ( f(x) ) over the interval ([5, 7]) is:

[ \frac{252 - 128}{7 - 5} = \frac{124}{2} = 62 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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