Can you help me find the critical numbers in this?

Firstly, I think what you meant was

f(t) = |3t-4| instead of f(x).

From there:

The domain of f(t) is the set of all real numbers.

Let us use the fact sqrt ((3t-4)^2) = |3t-4| to rewrite function f as follows:

f(t) = sqrt ((3t-4)^2)

Using chain rule with u-substitution (u(t) = 3t-4):

f'(t) = (1/2) 2 u(t) u'(t) / | u |

f'(t) = (3t-4)(3)/|3t-4|

Therefore, since f'(4/3) is undefined, 4/3 is the critical number (and the only critical number) for f(t)=|3t-4|.

Critical Numbers

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Critical points of a function occur where the function is not differentiable or where the derivative is zero.

Consider ƒ(t) =│3t-4│.

➊

Absolute value functions graph as a V and are, therefore, not differentiable at their vertex.

So the vertex is a critical point.

Since absolute value is always positive or zero, it follows that the minimum value of ƒ(t) is 0.

And, that is where the vertex is.

So, we need to solve:

f(t) =│3t-4│= 0

ƒ(t) = 0 for 3t-4 = 0; which is t = 4/3.

Therefore,

ƒ(t) has a critical point at t = 4/3.

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➋

To the left of t=4/3, the graph of f(t) =│3t-4│ is determined by the linear equation

ƒ(t) = -(3t-4) OR ƒ(t) = -3t+4. The derivative of this is the constant -3 so that to the left of 4/3 the derivative is never 0.

So, there are no critical points to the left of t=4/3.

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➌

Similarly, to the right of t=4/3, the graph of ƒ(t) is determined by the linear equation

ƒ(t) = 3t-4 which has as its derivative the constant 3; so that to the right of 4/3 the derivative is never 0.

So, there are no critical points to the right of t=4/3.

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From ➊, ➋, and ➌ we have that

ƒ(t) =│3t-4│ has only one critical point.

And, it is at t=4/3.

ANSWER

t = 4/3

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The absolute value is just a "smart -term" to make numbers positive.

Basically make the numbers within the absolute value, l...l. positive.

Step 1. Ignore the l...l for now

3t - 4 = 0

3t = 4

t = 4/3

Step 2. Make the values positive.

Since the answer is already positive, the is no need to make it positive again, hence it is 4/3

The critical number iswhere the function ceases to be differentiable. that is when t = 4/3

.

well have you ever thought of having two answers?

abs val usually means you are gonna have a +/- type thing

i would attempt to differentiate after you symplify to f(X)=3t-4 and f(x)=-3t+4

im not entirely sure but i think its 3/4 and 4