Answer:
(f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
Step-by-step explanation:
The function f⁻¹(x) is the reflection of the function f(x) across the line y=x. Every point (a, b) that is on the graph of f(x) is reflected to be a point (b, a) on the graph of f⁻¹(x).
Any line with slope m reflected across the line y=x will have slope 1/m. (x and y are interchanged, so m=∆y/∆x becomes ∆x/∆y=1/m) Since f'(x) is the slope of the tangent line at (x, f(x)), 1/f'(x) will be the slope of the tangent line at (f(x), x).
Replacing x with f⁻¹(x) in the above relation, you get ...
... (f⁻¹)'(x) = 1/f'(f⁻¹(x)) will be the slope at (x, f⁻¹(x))
Putting your given values in this relation, you get
... (f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
PROBLEM 9:
0.25
PROBLEM 8:
3/14
PROBLEM 7:
1 1/12 or 13/12
Answer:Assuming a 3-inch diameter, pillar candles of 3, 4 and 6 inches in height will burn for 45, 60 and 75 hours, respectively. Three-inch Party Lite pillar candles will burn 30-35 hours; 5-inch, 65 to 70 hours; and 7-inch, 90 to 95 hours. The three-wick 6-by-5-inch candles burn for more than 100 hours.
Step-by-step explanation:
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Answer:
10.6 feet
Step-by-step explanation:
Length = 22ft
width= 14ft
Maximum area = 800 sq. ft
Let X be increase in the number of feet for the length and width.
The new length = (22 + x) ft
New width= (14 + x) ft
Area = (22+x)(14+x) ≤ 800
308 + 36x + x^2 ≤ 800
x^2 + 36x + 308 - 800 ≤ 0
x^2 + 36x - 492 ≤ 0
Solve using quadratic equation
x = (-b +/- √b^2 - 4ac) / 2a
a= 1, b = 36, c= 492
x = (-36 +/- √36^2 - 4*1*-492)/ 2*1
= (-36 +/- √1396 + 1968) / 2
= (-36 +/- √3264) / 2
= (-36 +/- 57.13) / 2
x = (-36 + 57.13)/2 or (-36 - 57.13)/2
x = 21.13/2 or -93.13/2
x = 10.6 or -31.0
x = 10.6 ft
The length and width must increase by 10.6 ft each
