How do you find #(goh)(-4+a)# given #g(a)=2a+2# and #h(a)=-2a-5#?

Answer 1

To find ( (g \circ h)(-4+a) ), first, substitute ( -4+a ) into the function ( h(a) ) to find ( h(-4+a) ). Then, substitute the result into the function ( g(a) ) to find ( g(h(-4+a)) ).

Find ( h(-4+a) ): [ h(-4+a) = -2(-4+a) - 5 ]
Simplify: [ h(-4+a) = 8 - 2a - 5 ]

[ h(-4+a) = 3 - 2a ]

Now, find ( g(h(-4+a)) ): [ g(h(-4+a)) = 2(3 - 2a) + 2 ]
Simplify: [ g(h(-4+a)) = 6 - 4a + 2 ]

[ g(h(-4+a)) = 8 - 4a ]

So, ( (g \circ h)(-4+a) = 8 - 4a ).

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Answer 2

Genesis Allen

#8 - 4a#

Answer #x = g(h(a-4))#

#h(a - 4) = -2(a - 4) - 5 = -2a +8 - 5 = 3 - 2a#

Apply g.

#g(h(a - 4)) = g(3 - 2a) = 2(3 - 2a) + 2 = 6 - 4a + 2#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.