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abruzzese [7]
3 years ago
6

If ƒ(x) = 2x, then ƒ -1(x) = -2x ½x x - 2

Mathematics
1 answer:
kap26 [50]3 years ago
7 0

Answer: second option

Step-by-step explanation:

To find the inverse function of the given function f(x):

f(x)=2x

You need to:

Substitute f(x)=y into the function:

 y=2x

Now, you need  to solve for "x", to do this, you must divide both sides of the function by 2:

\frac{y}{2}=\frac{2x}{2}\\\\\frac{y}{2}=x

Now, replace "x" with "y" and replace "y" with "x":

\frac{x}{2}=y

Therefore, substituting y=f^{-1}(x) you get:

f^{-1}(x)=\frac{x}{2} or  f^{-1}(x)=\frac{1}{2}x

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If F(x)=x^3-2x^2, which expression is equivalent to f(I) -2+i,-2-i,2+i,2-i
Romashka-Z-Leto [24]

Hello from MrBillDoesMath!

Answer:

2 - i

Discussion:

Evaluate F(x) = x^3 - 2 x^2 when x = i.

i^2 = -1 for i^3 = i ( i^2) = i (-1) = -i

So F(i) =  (i)^3 - 2 (i)^2                      

          =   -i     - 2(-1)

          = -i      +  2

          =  2      -  i


which is the last answer shown


Thank you,

MrB

6 0
3 years ago
What is the value of log 6 36?
Zanzabum

Answer:

2

Step-by-step explanation:

Logarithm base 6 of 36 is 2 .

8 0
3 years ago
Help -------------------------
34kurt

Answer:

(A) -3 ≤ x ≤ 1

Step-by-step explanation:

The given function is presented as follows;

h(x) = x² - 1

From the given function, the coefficient of the quadratic term is positive, and therefore, the function is U shaped and has a minimum value, with the slope on the interval to the left of <em>h</em> having a negative rate of change;

The minimum value of h(x) is found as follows;

At the minimum of h(x), h'(x) = d(h(x)/dx = d(x² - 1)/dx = 2·x = 0

∴ x = 0/2 = 0 at the minimum

Therefore, the function is symmetrical about the point where x = 0

The average rate of change over an interval is given by the change in 'y' and x-values over the end-point in the interval, which is the slope of a straight line drawn between the points

The average rate of change will be negative where the y-value of the left boundary of the interval is higher than the y-value of the right boundary of the interval, such that the line formed by joining the endpoints of the interval slope downwards from left to right

The distance from the x-value of left boundary of the interval that would have a negative slope from x = 0 will be more than the distance of the x-value of the right boundary of the interval

Therefore, the interval over which <em>h</em> has a negative rate of change is -3 ≤ x ≤ 1

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