What must be added to x² + 6x²-x+ 5 to make it exact divisible by (x + 3)
1 answer:
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Answer:
The correct option is;
x ≥ 6; h⁻¹(x) = 6 +√(x - 22)
Step-by-step explanation:
Given the function, h(x) = x² - 12·x + 58
We can write, for simplification;
y = x² - 12·x + 58
Therefore;
When we put x as y to find the inverse in terms of x, we get;
x = y² - 12·y + 58
Which gives;
x - x = y² - 12·y + 58 - x
0 = y² - 12·y + (58 - x)
Solving the above equation with the quadratic formula, we get;
0 = y² - 12·y + (58 - x)
a = 1, b = -12, c = 58 - x
Therefore;
We note that for the function, h(x) = x² - 12·x + 58, has no real roots and the real minimum value of y is at x = 6, where y = 22 by differentiation as follows;
At minimum, h'(x) = 0 = 2·x - 12
x = 12/2 = 6
Therefore;
h(6) = 6² - 12×6 + 58 = 22
Which gives the coordinate of the minimum point as (6, 22) whereby the minimum value of y = 22 which gives √(x - 22) is always increasing
Therefore, for x ≥ 6, y or h⁻¹(x) = 6 +√(x - 22) and not 6 -√(x - 22) because 6 -√(x - 22) is less than 6
The correct option is x ≥ 6; h⁻¹(x) = 6 +√(x - 22).
N.b., this is copied from an answer I wrote for someone else.
Thinking about the graph of y, the rate of change is zero whenever there is an extremum.
First, differentiate y.
Next, find the zeros of y' by factoring.
Now, we substitute these x-values into the original equation to find the coordinates.
Thinking about the graph of y, the rate of change is zero whenever there is an extremum.
First, differentiate y.
Next, find the zeros of y' by factoring.
Now, we substitute these x-values into the original equation to find the coordinates.