Round the number 157 to the nearest tens plce
1 answer:
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Answer:
(a) -69.59 . . . . the approximate slope of the curve at x=2
(b) 26.583 . . . . the value of x where the slope of f(x) is about -25.
Step-by-step explanation:
(a) The slope of a function <em>f </em>over the interval [x, x+h] is given by ...
... slope = (f(x+h) -f(x))/h
The problem statement asks for the value of ...
... (f(2.01) -f(2))/0.01
Matching this expression to the pattern above, we find x = 2, h = 0.01. An appropriate interpretation of the expression is that it gives the approximate slope of f(x) near x=2.
The value of that slope is about ...
... (f(2.01) -f(2))/0.01 ≈ 1850·(0.92122386 -0.9216)/.01 ≈ -69.5857
(b) f(x+1) = 1850·0.96^(x+1) = 0.96·f(x)
Then the equation becomes ...
... 0.96·f(x) = f(x) -25
... 25 = 0.04·f(x) . . . . . . . . add 25-0.96·f(x)
... 625 = f(x) . . . . . . . . . . . multiply by 25
... 625 = 1850·0.96^x . . . substitute definition of f(x)
... 625/1850 = 0.96^x . . . divide by the coefficient of the exponential term
... log(625/1850) = x·log(0.96) . . . . take logs
... x = log(625/1850)/log(0.96) ≈ -0.471292/-0.0177288
... x ≈ 26.583
If you rearrange the equation to ...
... f(x+1) -f(x) = -25
Comparing this to the equation for slope, above, it looks a lot like we're trying to find the interval [x, x+1] over which the average slope is -25.
