The temperature in an industrial pasteurization tank is f (x) = x^2 - 7x + 111 d

Question

The temperature in an industrial pasteurization tank is f (x) = x^2 - 7x + 111 degrees centigrade after x minutes (for 0 lessthanorequalto x lessthanorequalto 12). (a) Find f' (x) by using the definition of the derivative. f' (x) = (b) Use your answer to part (a) to find the instantaneous rate of change of the temperature after 2 minutes. Be sure to interpret the sign of your answer. The temperature is at a rate of degrees per minute after 2 minutes. (c) Use your answer to part (a) to find the instantaneous rate of change after 9 minutes. The temperature is at a rate of degrees per minute after 9 minutes. For the function, find and simplify f (x + h) - f (x)/h (Assume h notequalto 0.) f (x) = 5x^2 - 2x + 3 f (x + h) - f (x)/h = 5h + 6x + 4

Explanation / Answer

7.

A.

f(x) = x^2 - 7x + 111

f'(x) = d(f(x))/dx = d(x^2 - 7x + 111)/dx

f'(x) = 2x - 7 + 0

f'(x) = 2x - 7

B.

at t = 2 min

f'(x) = 2*2 - 7

f'(x) = -3 deg/min

since f'(x) is negative temperature is decreasing at a rate of 3 deg/min.

C.

at t = 9 min

f'(x) = 2*9 - 7 = +11 deg/min

since f'(x) is positive, temperature is increasing at a rate of 11 deg per min

7.

f(x+h) = 5*(x + h)^2 - 2*(x + h) + 3

f(x) = 5x^2 - 2x + 3

[f(x + h) - f(x)]/h = [(5x^2 + 5h^2 + 10xh - 2x - 2h + 3) - (5x^2 - 2x + 3)]/h

= (5h^2 + 10xh - 2h)/h

[f(x + h) - f(x)]/h = 5h + 10x - 2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.