concavity quiz
Let f be a function with first derivative given by f′(x)=x(x−5)2(x+1). At what values of x does f have a relative maximum?
-1 only
Let f be the function given by f(x)=5cos2(x2)+ln(x+1)−3. The derivative of f is given by f′(x)=−5cos(x2)sin(x2)+1x+1. What value of c satisfies the conclusion of the Mean Value Theorem applied to f on the interval [1,4] ?
2.749 because f′(2.749)=(f(4)−f(1))/3
Selected values of a continuous function f are given in the table above. Which of the following statements could be false?
By the Mean Value Theorem applied to f on the interval [2,5], there is a value c such that f′(c)=10.
Let f be the function given by f(x)=x(x−4)(x+2) on the closed interval [−7,7]. Of the following intervals, on which can the Mean Value Theorem be applied to f ?
I and II only
The temperature inside a vehicle is modeled by the function f, where f(t) is measured in degrees Fahrenheit and t is measured in minutes. The first derivative of f is given by f′(t)=t2−3t+cost. At what times t, for 0<t<4, does the temperature attain a local minimum?
3.299
Let f be the function given by f(x)=x+4(x−1)(x+3) on the closed interval [−5,5]. On which of the following closed intervals is the function f guaranteed by the Extreme Value Theorem to have an absolute maximum and an absolute minimum?
[-2,0]
The derivative of the function f is given by f′(x)=x2−2−3xcosx. On which of the following intervals in [−4,3] is f decreasing?
[−3.444,−1.806] and [−0.660,1.509]
Let f be the function defined by f(x)=xlnx for x>0. On what open interval is f decreasing?
0<x< 1/e only
The graph of f′, the derivative of the function f, is shown above for 0<x<9. Which of the following statements is true for 0<x<9 ?
f has one relative minimum and two relative maxima
Let f be the function defined by f(x)=x3−6x2+9x+4 for 0<x<3. Which of the following statements is true?
f is decreasing on the interval (1,3) because f'(x)<0 on the interval (1,3)
Let f be a differentiable function with f(0)=−4 and f(10)=11. Which of the following must be true for some c in the interval (0,10) ?
f′(c)=(11−(−4))/(10−0), since the Mean Value Theorem applies.
Let f be the function defined by f(x)=xsinx with domain [0,∞). The function f has no absolute minimum and no absolute maximum on its domain. Why does this not contradict the Extreme Value Theorem?
the domain of f is not a closed and bounded interval