Given that the function f(x)=1/2x-6 was replaced with f(x+k)=1/2x-4, then the value of k we be as follows:
f(x)=1/2x-6
replacing x1=x+k
f(x+k)=1/2(x+k)-6
equating the above to 1/2x-4 we shall have:
1/2(x+k)-6=1/2x-4
simplifying we obtain:
1/2x+1/2k=1/2x-4
1/2k=-4
hence
k=-8
The answer to this question is:
671.44
Critical points: (Maximum, minimum, and inflection points) have a slope = 0
The derivative gives the slope, take the derivative set it equal to zero
<span>f(x) = x^5 -10x^3 + 9x
derivative:
f'(x) = 5x^4 - 30x^2 + 9
0 = 5x^4 - 30x^2 + 9
use quadratic formula (or polysolver) with a=5, b=-30 and c = 9
4th degree gives 4 solutions
x = +/- √(3 - 6/√(5))
x ~ +/- 2.38
x = +/- √(3 - 6/√(5))
x ~ +/- 0.563
</span>Determine which point is the highest (maximum) and the lowest (minimum) by putting the x-coordinates back into the original equation and comparing y-values.
* You will also need to check the endpoints given as these can sometime be the max/min of the function.
Our x-values:
x = { -3, -2.38, -0.563, 0.563, 2.38, 3}
respective y-values found by inputting x's into function
y = {0, 37.03, -3.34, 3.34, -37.03, 0}
nice symmetry.. check out the graph to see the solutions are correct.
https://www.desmos.com/calculator
Maximum point (-2.38, 37.03)
Minimum point (2.38, -37.03)
Please specify your question and give in detail..