Given that the f(y) = y - 1 and g(y) = y^3 - 2y - 1, find (gof) (y)I and hence, (gof) (-6).

1 answer:

3 0

$(g \circ f)(y)=(y-1)^3-2(y-1)-1\\(g \circ f)(y)=y^3 - 3 y^2 + 3 y - 1-2y+2-1\\(g \circ f)(y)=y^3 - 3 y^2 +y\\\\(g \circ f)(-6)=(-6)^3-3\cdot(-6)^2+(-6)\\(g \circ f)(-6)=-216-108-6\\(g \circ f)(-6)=-330$

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You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.

postnew [5]

This problem seems to be wrong because no minimum point was found and no point of landing exists

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

Derivatives

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

The function provided in the question is

$h(x)=2x^2-12x+20$

Let's find the first derivative

$h'(x)=4x-12$

solving h'=0:

$4x-12=0$

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

To find when will the ball land, we set h=0

$2x^2-12x+20=0$