Answer:
Range: 1 > x > 3
Step-by-step explanation:
We are given then function;
y = x³ - 6x² + 9x + 5
dy/dx = 3x² - 12x + 9
We want to find the range of values of x for which dy/dx > 0
Thus;
3x² - 12x + 9 > 0
Using quadratic formula, we have;
x = 1 and x = 3
Let's define the range;
try x = 0
3(0²) - 12(0) + 9 = 9
This is more than 0, so it is part of the solution.
Let's try x = 4
3(4²) - 12(4) + 9 = 9
This is more than 0 so it is part of the solution
Let's try x = 2
3(2²) - 12(2) + 9 = -3
This is less than 9 and so it's not part of the solution.
Thus the range of values of x for which dy/dx > 0 is;
1 > x > 3
-9x+5<17 answer x >-4/3 how I do to it
subtract -5 from both side and you get 12. so it be -9x<12
than you divide -9 from both side because you want x by it self you get -4/3 but you have to flip the sign since is negavite and you will be left over with ×>-4/3
<span>Part a) the max of g(t) occurs when t=0.6 seconds, for f(t) since the parabola is concave down due to the negative sign then we know that there is a max on the parabola, the max for f(t) occurs at (13/16,169/16) 169/6=10.5625inches at time 0.8125seconds
Part b) f(t) has greater x intercepts, meaning that when the labrador will land later than the fox when it jumps over the hurdle</span>