Explain how to find (f^-1)’(b), given that b=f(a).

(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)

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58