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)
I believe that 30.5 as a fraction is 30 and 1 half but don't take my word
The length of the line AB as described by points A and B in the task content is; √125.
<h3>What is the length of the line?</h3>
a) The length of AB as described is;
AB = √((7-2)² + (4-(-6))²)
AB = √(25 + 100) = √125.
b) The slope of AB is;
Slope, m = (7-2)/(4-(-6)) = 5/10
Slope = 1/2.
c). The coordinates of the midpoint of AB are;
x = (4+(-6))/2 = -1
y = (7+2)/2 = 4.5
The midpoint is; (-1, 4.5).
2) The radius of the circle as described with centre (0, 0) that passes through the point (5, –3) is;
Radius = √((5-0)² + (-3-0)²)
Radius = √(25+9) = √34 = 5.8
3) Hence, the equation of a circle with centre (0, 0) that passes through the point (5, –3) is;
(x-5)² + (x-3)² = 5.8².
Read more on length of a line;
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X=2 it’s Honestly easy I just look at da missing number