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3 Answers
- MyRankLv 62 years ago
f(x) = x³ + 4x + 3
f’(x) = 3x² + 4
f(x) = y → x = f¯¹(y)
y = 3x² + 4
x² = y - 4 / 3
= √y - 4/3
f¯¹(g) = √y - 4/3
f¯¹(3) = √3 - 4 / 3
= √-1/3.
Source(s): http://myrank.co.in/ - 2 years ago
Let's look at another example of a function, it's inverse function, and their derivatives to see how they relate
f(x) = x^3
f'(x) = 3x^2
f-1(x) = x^(1/3)
f-1(x) = (1/3) * x^(-2/3)
f(2) = 2^3 = 8
f'(2) = 3 * 2^2 = 3 * 4 = 12
f-1(8) = 8^(1/3) = 2
f-1'(8) = (1/3) * 8^(-2/3) = (1/3) * (1/4) = 1/12
So the relationship is hopefully clear.
f(x) = x^3 + 4x + 3
f-1(3) = ?
f-1(3) = k
3 = f(k)
f(k) = k^3 + 4k + 3
3 = k^3 + 4k + 3
0 = k^3 + 4k
0 = k * (k^2 + 4)
k = 0 , -2i , 2i
k = 0 is the only real value
f(x) = x^3 + 4x + 3
y = x^3 + 4x + 3
x = y^3 + 4y + 3
1 = 3y^2 * y' + 4 * y'
1 = (3y^2 + 4) * y'
y' = 1 / (3y^2 + 4)
f(8) = 3
f-1(3) = 8
y' = 1 / (3 * 8^2 + 4) = 1 / (3 * 64 + 4) = 1 / (192 + 4) = 1/196
f(x) = x^3 + 4x + 3
f'(x) = 3x^2 + 4
f'(8) = 3 * 8^2 + 4 = 3 * 64 + 4 = 192 + 4 = 196
f-1'(3) = 1/f'(8) = 1/196
