If f(x) is continuous and differentiable and #f(x) = ax^4 + 5x#; #x<=2# and #bx^2 - 3x#; x> 2, then how do you find b?

Answer 1

#b = 15/2#

As #f(x)# is continuous at #x=2#, we have
#lim_(x->2^-)f(x) = lim_(x->2^+)f(x)#
#=> a(2^4)+5(2) = b(2^2)-3(2)#
#=> 16a+10 = 4b-6#
#=> a = 1/4b - 1#
As #f(x)# is differentiable at #x=2#, the limit #f'(2) = lim_(x->2)(f(x)-f(2))/(x-2)# must exist. We can tell what the one sided limits will evaluate to by calculating the derivatives of the components of the piecewise defined functions on either side of #2#.
#lim_(x->2^-)(f(x)-f(2))/(x-2) = lim_(x->2^+)(f(x)-f(2))/(x-2)#
#=> 4a(2^3)+5 = 2b(2)-3#
#=> 32a+5 = 4b-3#
Substituting in #a = 1/4b-1#, we have
#32(1/4b-1) + 5 = 4b-3#
#=> 8b - 27 = 4b - 3#
#=> 4b = 30#
#:. b = 15/2#
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Answer 2

To find the value of (b), we need to ensure that the function (f(x)) is continuous at (x = 2). For a function to be continuous at a point, the left-hand limit and the right-hand limit must be equal, and they must equal the value of the function at that point.

Given (f(x) = ax^4 + 5x) for (x \leq 2) and (f(x) = bx^2 - 3x) for (x > 2), we can set up the continuity condition:

[f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)]

First, let's find (f(2)): [f(2) = a(2)^4 + 5(2) = 16a + 10]

Now, let's find the left-hand limit as (x) approaches (2): [\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (ax^4 + 5x) = a(2)^4 + 5(2) = 16a + 10]

And the right-hand limit as (x) approaches (2): [\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (bx^2 - 3x) = b(2)^2 - 3(2) = 4b - 6]

For (f(x)) to be continuous at (x = 2), the left-hand limit must equal the right-hand limit, so:

[16a + 10 = 4b - 6]

We can then solve this equation for (b): [4b = 16a + 10 + 6] [4b = 16a + 16] [b = 4a + 4]

So, the value of (b) is (4a + 4).

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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.

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.

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.

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.

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