111q1 Winter 2020 solution PDF

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MATH 111–002 Winter 2020 Solution to Quiz 1 Date: January 20, 2020 Time: 25 minutes

Total marks: 20

1. (4 marks) Show that the function f (x) = 2x9 + 8x − π is invertible. Solution.

• The domain of f is the interval R = (−∞, ∞).

• The derivative is f ′ (x) = 18x8 + 8. Since f ′ (x) > 0 for every x ∈ R, it follows that f is increasing everywhere.

Combining these two items, we get that f is invertible. 2. (4 marks) Let f (x) =

2x−1 . 3x+2

Find a formula for the inverse of f (x).

Solution. • Write y =

2x−1 . 3x+2

• Solving this equation for x: multiplying both sides by (3x+ 2), we get y(3x+ 2) = 2x − 1, which is equivalent to 3yx + 2y = 2x − 1. Gathering the x’s on one side and the y’s on the other side, we get 3yx − 2x = −2y − 1 or x(3y − 2) = −2y − 1. Dividing both sides by 3y − 2, we get x = −2y−1 . 3y−2

• Swapping x and y, we get y =

−2x−1 . 3x−2

So the inverse function is f −1 (x) =

−2x − 1 3x − 2

or

f −1 (x) =

2x + 1 . −3x + 2

3. (3 marks) Let y = f (x) be a continuous, differentiable, and invertible function. Suppose f (0) = 2 f (2) = −3 f (−3) = 0 ′ f (0) = 1 f ′ (−3) = −2 f ′ (2) = 5 Find f −1 (0) and (f −1 )′ (−3). Solution. We have

f −1 (0) = −3 and

1 1 1 = ′ = . f ′ (f −1 (−3)) f (2) 5 √ 4. (3 marks) Sketch the graphs of f (x) = x + 1 and its inverse function. (f −1 )′ (−3) =

Solution. The graphs of f and f −1 are shown in Figure 1.

1

Figure 1: Graph of y = x3 + 2 5. (3 marks) Find the derivative of f (x) = (sin x)e2x . Using the product rule (and then the chain rule), we get f ′ (x) =

d d d [sin x]e2x + (sin x) [e2x ] = (cos x)e2x + (sin x)e2x [2x] = dx dx dx

(cos x)e2x + (sin x)e2x (2) = e2x (cos x + 2 sin x). R 2 6. (3 marks) Find the integral xex dx. Solution.

• Let u = x2 •

du dx

= 2x, so that dx =

du . 2x

• The integral becomes Z Z Z 1 1 1 2 x2 u du xe dx = xe = eu du = eu + C = ex + C. 2 2x 2 2

2...