Unit 2 Progress Check: MCQ Part A

If f is the function defined by f(x)=x√4, what is f′(x)?

(1/4)x^(−3/4) f(x)= 4th√x = x^(1/4) is a power function. Therefore, the power rule is applied to determine the derivative. f′(x)=ⅆⅆx(x^(1/4))=(1/4)x^(−3/4)

If f(x)=x^5, then f′(x)=

5x^4 According to the power rule for differentiation, the derivative of functions of the form f(x)=x^r is f′(x)=rx^(r−1).

Selected values of a function f are shown in the table above. What is the average rate of change of f over the interval [1,5] ?

(14-2)/(5-1) The average rate of change over the interval [1,5][1,5] is given by the difference quotient [f(5)−f(1)]/(5−1) = (14−2)/(5−1).

The function f is given by f(x)=1+3cosx. What is the average rate of change of f over the interval [0,π] ?

-6/π The difference quotient [f(π)−f(0)]/(π−0) is the average rate of change of ff over the interval [0,π][0,π].

If f(x)=1/(x^7), then f′(x)=

-7/(x^8) Writing 1/(x^7) as x^(−7), the power rule for differentiation gives f′(x)=−7x^−8=−7x^8 f′(x)=−7x−8=−7/x^8.

The graph of the function f, shown above, consists of three line segments. What is the average rate of change of f over the interval −1≤x≤6 ?

0 The average rate of change of ff over the interval [−1,6][−1,6] is given by the difference quotient [f(6)−f(−1)]/(6−(−1)) = (0−0)/7 = 0.

Let f be the function given by f(x)=2x3. Selected values of f are given in the table above. If the values in the table are used to approximate f′(0.5), what is the difference between the approximation and the actual value of f′(0.5) ?

0.433 The numerical value of the derivative at x=0.5 obtained from the calculator is f′(0.5)=0.567. A difference quotient can be used with the values in the table to estimate the derivative as [f(1)−f(0)]/(1−0) = (2−1)/1 = 1. The error between the actual derivative value and this approximation is 0.433.

The derivative of a function f is given by f′(x)=0.1x+e^0.25x. At what value of x for x>0 does the line tangent to the graph of f at x have slope 2 ?

2.287 Since the derivative at a point is the slope of the line tangent to the graph at that point, the calculator is used to solve f′(x)=0.1x+e^(0.25x)=2.

The graph of the trigonometric function f is shown above for a≤x≤b. At which of the following points on the graph of f could the instantaneous rate of change of f equal the average rate of change of f on the interval [a,b] ?

B The instantaneous rate of change of ff at the point BB is the slope of the line tangent to the graph of ff at the point BB. The average rate of change of ff on the interval [a,b][a,b] is the slope of the secant line through the points (a,f(a))(a,f(a)) and (b,f(b))(b,f(b)). The tangent line at BB appears to be parallel to the secant line. Therefore, the instantaneous rate of change at BB could be equal to the average rate of change.

Which of the following statements, if true, can be used to conclude that f(2) exists? i. limx→2f(x) exists. ii. f is continuous at x=2. iii. f is differentiable at x=2.

II and III only If ff is continuous at x=2, then f(2) exists. Also if f is differentiable at x=2, then ff is continuous at x=2 and f(2) exists.

f(x)={3x+15x−3forx≤2forx>2 Let f be the function defined above. Which of the following statements is true?

f is continuous but not differentiable at x=2 ff is continuous at x=2 because [skipped work] f(2)=7.f is not differentiable at x=2 because the left-hand derivative does not equal the right-hand derivative at x=2, as follows.

The graph of the function f, shown above, has a vertical tangent at x=−2 and horizontal tangents at x=−3 and x=−1. Which of the following statements is false?

f is not differentiable at x=−3 and x=−1 because the graph of ff has horizontal tangents at x=−3 and x=−1. A function is differentiable at points at which the line tangent to the graph is horizontal.

Let f be the function given by f(x)=(1/7)x^7+12x^6−x^5−(15/4)x^4+(4/3)x^3+6x^2. Which of the following statements is true?

f′(0.4)<f′(−1.5)<f′(−3.1)f′(0.4)<f′(−1.5)<f′(−3.1) The calculator is used to store the expression for f(x) and to find the numerical derivative values at the three values of xx. They are, in increasing order, f′(0.4)=4.387, f′(−1.5)=4.922, and f′(−3.1)=14.974.

The graph of f′, the derivative of a function f, is shown above. The points (2,6) and (4,18) are on the graph of f. Which of the following is an equation of the line tangent to the graph of f at x=2 ?

y=5x−4 The slope of the line tangent to the graph of ff at x=2 is f′(2). This value can be read from the graph of f′and is equal to 5. An equation of the line with slope 5 that contains the point (2,6) is y=5(x−2)+6=5x−4.

The derivative of the function f is given by f′(x)=−3x+4 for all x, and f(−1)=6. Which of the following is an equation of the line tangent to the graph of f at x=−1 ?

y=7x+13 The slope of the line tangent to the graph of ff at x=−1 is f′(−1)=−3(−1)+4=7. An equation of the line containing the point (−1,6) with slope 7 is y=7(x+1)+6=7x+13