Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
If <em>h(x)</em> is the inverse of <em>f(x)</em>, then by definition of inverse function, we have
<em>f(h(x))</em> = <em>x</em>
By definition of <em>f(x)</em>, composing <em>f</em> with <em>h</em> gives
<em>f(h(x))</em> = 2 <em>h(x)</em> - 10
Solve for <em>h(x)</em> :
<em>x</em> = 2 <em>h(x)</em> - 10
2 <em>h(x)</em> = <em>x</em> + 10
<em>h(x)</em> = (<em>x</em> + 10)/2
<em>h(x)</em> = <em>x</em>/2 + 5
So... whichever option most closely resembles this is the correct one. It's hard to tell exactly which one that would be.
Answer:
In the picture.
Step-by-step explanation:
I hope that it's a clear solution.
I think yes because they are both the same (or congruent) because they both equal 180°