Find the critical numbers of the function g(t)=|5t-7|
If you find g'(t) you just get 5 so would there be no critical numbers? Or am I missing something?
your graph consists of two straight lines
y = 5t-7 and y = -5t + 7 where you only want the part above the t-axis (x-axis normally)
The two lines meet on the t-axis at the point (7/5 , 0)
That point is called a "cusp" and it would be the only critical number.
Differentiating the function would be over-kill since when you find the derivative of a straight line, you just get the slope, which we knew already anyway.
To find the critical numbers of the function g(t) = |5t - 7|, we need to determine the values of t at which the derivative g'(t) does not exist or is equal to zero.
Let's start by finding the derivative of g(t):
g'(t) = d/dt |5t - 7|
To find the derivative of the absolute value function, we need to consider two cases: when the argument (5t - 7) is positive and when it is negative.
Case 1: (5t - 7) > 0
In this case, the absolute value is simply the argument itself, so the derivative is:
g'(t) = d/dt (5t - 7) = 5
Case 2: (5t - 7) < 0
Now, the absolute value is the argument with a negative sign, so the derivative is:
g'(t) = d/dt (-(5t - 7)) = -5
As you correctly pointed out, the derivative g'(t) of the absolute value function is either 5 or -5, depending on the case. It does not depend on the value of t.
Therefore, g'(t) does not equal zero at any value of t, since it is always either 5 or -5. Consequently, there are no critical numbers for the function g(t)=|5t-7|.