Find the rate of change of the function h(x) = 4x – 3 on the interval 0 ≤ x ≤ 1. The rate of change is?

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Find the rate of change of the function h(x) = 4x – 3 on the interval 0 ≤ x ≤ 1. The rate of change is?

Mathematics
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rate of change from x = A to x = B for h(x) is [h(B)-h(A)]/(B-A) let B = 1 and A = 0, then evaluate the expression
how do I set this up?
[h(1)-h(0)]/(1-0)

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Other answers:

first fing came to my was derivative but it doesn't need it ;)
[h(1)-h(0)]/(1-0)
was that the answer?
no, you still need to plug in h(1), h(0), and simplify the denominator...
what is h(1)?
idk im lost please explain :(
h(x) = 4x-3 so what is h(1)?
4?
no, please try again h(1) means let x = 1 and calculate h(x) if h(x) = 4x - 3 and x = 1 then what is h(x)?
idk
h(x) = 4x - 3 and x = 1 replace "x" with "1" and solve h(x) = 4x -3
so whats the answer and then explain please
x = 1 so what is 4x - 3?
the answer is 1?
right, 4x - 3 = 1 when x = 1 BUT we are not done yet h(1) = 1, since h(x) = 1 when x = 1 now, we need to find h(0) if x = 0, and h(x) = 4x - 3, then what is h(x)?
4
no, please try again... x = 0 4x - 3 = ?
1
no, one more time please x = 0 4x - 3 = 4*(0) - 3 = ?
3
what is 4*0?
0
yes, so what is 4*0 - 3?
-3
good, so we know that h(0) = -3 and h(1) = 1 so, what is [h(1)-h(0)]/(1-0)?
i cant read that
|dw:1436822012814:dw| better?
evaluate the numerator first, then the denominator, then tell me what you get
1 over 1
not quite, try the numerator again... h(1) = 1 h(0) = -3 h(1)-h(0) = ?
idk
we know that h(1) = 1 and h(0) = -3 replace "h(1)" with 1 and "h(0)" with -3 what is h(1)-h(0)?
|dw:1436822235650:dw|
-3
|dw:1436822597360:dw|
what is 1-(-3)?
hey buddy whats thw anser please
then explain it
because i am really lost right now
|dw:1436823130155:dw|
our numerator is h(1) - h(0) h(1) = 1 h(0) = -3 so h(1) - h(0) = 1 - (-3) = ?
-2
1 - (-3) = 1 + 3 = ?
4
right, so we know that h(1) - h(0) = 4 now, can you tell me what this equals? |dw:1436823630938:dw|
equals -3
not quite, please try one more time... |dw:1436824260512:dw|

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