help with pre-calculus and calculus simple question but i dont remember how to do?

Question

instructions are....
assume it has an inverse find the func. values..dont dont solve fer the inverse
problem is..x3rd+4x-1
a)f negative 1(-1)
b)fnegative1 (4)
explain how you got the answer and another problem too
14. x3rd+2x+1
a)f-1 (1)
b)f-1 (13)

Answer 1

HA! I actually tried to solve for the inverse. I see why the problem told you NOT to do so. That's a toughie.

But finding the f­­­­­­­­ˉ­¹(x) values is very simple.

Just remember that the if ordered pair (x,y) is in f(x), then the ordered pair (y,x) is in the inverse f­­­­­­­­ˉ­¹(x). The x and y-values switch places -- that is all an inverse is.

You want f­­­­­­­­ˉ­¹(-1) and f­­­­­­­­ˉ­¹(4).
Just plug f(x), the original equation
f(x) = x³ + 4x -1
into your calculator and look at the table of values. No need to graph it, you just need the table.

The x-values that correspond to the y-values -1 and 4 are the answers.

For the given function, f(x), I see the ordered pairs (0, -1) and (1,4).

This means that for the inverse function f­­­­­­­­ˉ­¹(x), the ordered pairs are (-1, 0) and (4, 1). Your answers are 0 and 1, respectively.

Answer 2

a) x^3 + 4x - 1 = - 1
The answer matches the constant, so
x^3 + 4x = 0
x(x^2 + 4) = 0
f^-1(-1) = 0, i2, - i2
b) x^3 + 4x - 1 = 4
x^3 + 4x - 5 = 0
Find x = 1 by guess & check.
(x - 1)(x^2 + x + 5) = 0
x = 1, (1/2)(1 + i√19), (1/2)(1 - i√19)
f^-1(4) = 1, (1/2)(1 + i√19), (1/2)(1 - i√19)

f(x) = x^3 + 2x + 1
a) f^-1(1) = 0, i√2, - i√2
x^3 + 2x + 1 = 13
x^3 + 2x - 12 = 0
(x - 2)(x^2 + 2x - 6) = 0
x = 2, 1 + √6, 1 - √6
b) f^-1(13) = 2, 1 + √6, 1 - √6

Answer 3

I consider that to uncover the inflection aspects, you must take the moment spinoff of the equations after which set them same to 0 and clear up for x. So... f(x) = x^two - 2x + one million f'(x) = 2x - two f''(x) = two and g(x) = x^three - 3x^two + two g'(x) = 3x^two - 6x g''(x) = 6x - 6 6x-6 = zero x=one million (one million, zero)