# Let f(x) = x3 – 7×2 + 25x – 39 and let g be the inverse function of f. What is the value of g'(0)=?

0

a) -1/25

b) 1/25

c) 1/10

d) 10

e) 25

• f(x) = x^3 – 7x^2 + 25x – 39

f'(x) = 3x^2 – 14x + 25

Since g is the inverse function of f

f(g(x)) = x

Differentiate both sides of equation

f'(g(x)) g'(x) = 1

Now f(3) = 0, so g(0) = 3

g'(x) = 1 / f'(g(x))

g'(0) = 1 / f'(g(0))

g'(0) = 1 / f'(3)

g'(0) = 1 / 10

• f(x) = x^3 – 7x^2 + 25x – 39 Inverse x = g(x)^3 – 7g(x)^2 +25g(x) – 39 discover first derivative a million = 3g(x)^2*g'(x) – 14g(x)*g'(x) + 25g'(x) utilising ability rule and chain rule clean up for g'(x) g'(x) = a million /(3g(x)^2 – 14g(x) + 25) Plug in 0 for x g'(0) = a million/(3g(0)^2 – 14g(0) + 25) bear in mind inverse & plug x = 0 x = g(x)^3 – 7g(x)^2 +25g(x) – 39 0 = g(0)^3 – 7g(0)^2 + 25g(0) – 39 Yuck! feels like we could factor this. because of the fact the final coefficient is 39, recommendations to objective are a million, 3, 13, or 39 So i’m getting: 0 = (g(0) – 3)(g(0)^2 – 4g(0) + 13) so g(0) – 3 = 0 and g(0)^2 – 4g(0) + 13 = 0 so g(0) = 3 and no real recommendations for the 2nd equation because of the fact the discriminate, b^2 – 4ac, is decrease than 0. So the only real answer is g(0) = 3 Now plug 3 in for g(0) in g'(0) above. . g'(0) = a million/(3*3^2 – 14*3 + 25) which simplifies to a million/10.

• f(x) = x^3 – 7x^2 + 25x – 39

f'(x) = 3x^2 – 14x + 25

Since g is the inverse function of f

f(g(x)) = x

Differentiate both sides of equation

f'(g(x)) g'(x) = 1

Now f(3) = 0, so g(0) = 3

g'(x) = 1 / f'(g(x))

g'(0) = 1 / f'(g(0))

g'(0) = 1 / f'(3)

g'(0) = 1 / 10

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