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Consider the following function. h(t) = t/t-5

h '(t) = Find the derivative of the function.
Find all the values of t for which
h '(t) = 0
or
h '(t) is discontinuous.

User Sgmoore
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Final answer:

To find the derivative of the function h(t) = t/(t-5), you can use the quotient rule. The derivative is -5/((t-5))^2. There are no values of t for which h'(t) = 0, and h'(t) is continuous for all values of t except t = 5.

Step-by-step explanation:

To find the derivative of the function h(t) = t/(t-5), we can use the quotient rule. The quotient rule states that if we have a function of the form f(t)/g(t), the derivative is given by (f'(t) * g(t) - f(t) * g'(t))/(g(t))^2.

Applying this rule to h(t), we have h'(t) = (1 * (t-5) - t * 1)/((t-5))^2.

Simplifying further, h'(t) = -5/((t-5))^2.

To find the values of t for which h'(t) = 0, we can set -5/((t-5))^2 equal to 0. This will occur when the numerator, -5, is equal to 0. However, since -5 is a constant, it can never be equal to 0. Therefore, there are no values of t for which h'(t) = 0.

In terms of discontinuity, the function h(t) = t/(t-5) is continuous for all values of t except t = 5. At t = 5, the denominator becomes zero, which causes the function to become undefined. Therefore, h'(t) is continuous for all values of t except t = 5.

User MrZiggyStardust
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