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PRACTICE PROBLEMS: SET 3 MATH 100: PROF. DRAGOS GHIOCA

1. Problems   11π  Problem 1. Find arcsin sin 3 .

Problem 2. Find the derivative of the functions: (a) y(x) = 53

x

(b) y(x) = sin(x)ln(x) Problem 3. Find the inverse function for x

f (x) = 53 . 2

Problem 4. Find the inverse function for f (x) = 3x where f : [0, +∞) −→ [1, +∞). Problem 5. (1) If f (x) = ex +xe for positive real numbers x, and g(x) is the inverse function for f (x), find g (2ee ). (2) If f (x) = (x2 + 1)x for positive real numbers x, and g(x) is the inverse of f (x), find g(2).

2. Solutions.   11π  Problem 1. arcsin sin 3 is an angle θ with the properties: π π • − 2 ≤ θ ≤ 2 ; and  . • sin(θ) = sin 11π 3 We know that sin(x) is a function periodic of period 2π; so,         11π 11π −π 11π − 2π = sin − 4π = sin . = sin sin 3 3 3 3 or 11π − 2π = So, θ = − π3 . Note that 11π 3  3 π π are not in the interval − 2 , 2 . Problem 2.

1

5π 3

are not the correct answer since they

2

MATH 100: PROF. DRAGOS GHIOCA

(a) We can either differentiate directly and get x

x

y′ (x) = 53 · ln(5) · (3x ) = 53 · ln(5) · 3x · ln(3). ′

Alternatively we can use logarithmic differentiation and get ln(y(x)) = 3x · ln(5) and then we differentiate (note the chain rule for the left hand side) y′ (x) = 3x · ln(3) · ln(5), y(x) and so x

y′ (x) = y(x) · 3x · ln(3) · ln(5) = 53 · 3x · ln(3) · ln(5). (b) We use logarithmic differentiation: ln(y(x)) = ln(x) · ln(sin(x)), and then differentiate 1 cos(x) y′ (x) , = · ln(sin(x)) + ln(x) · sin(x) x y(x) and so ′

ln(x)

y (x) = sin(x)

·



ln(x) · cos(x) ln(sin(x)) + x sin(x)



.

Problem 3. We let y = g(x) be the inverse function of f (x). Then f (g(x)) = x, or in other words, f (y) = x. But f (y) = 53 ; so y

y

53 = x. We apply the logarithm in base 5 to both sides and compute 3y = log5 (x). Then we apply the logarithm in base 3 and get y = g(x) = log3 (log5 (x)) is the inverse function for f (x). Second solution: In order to find y = g(x) the inverse function of f (x), we could work only with the natural logarithm ln. So, from f (g(x)) = x which yields f (y) = x and thus y 53 = x. We apply the natural logarithm and get 3y · ln(5) = ln(x), and so, ln(x) . ln(5) Then applying again the natural logarithm yields   ln(x) , y · ln(3) = ln ln(5) 3y =

PRACTICE PROBLEMS: SET 3

3

and thus the inverse function is ln y = g(x) =



ln(x) ln(5)

ln(3)



.

The two expressions for g(x) computed with the above two methods are the same since ln(x) log5 (x) = . ln(5) Similarly, for any t we have ln(t) log3 (t) = . ln(3) The above properties of the logarithm are just as important as the identities: loga (A · B) = loga (A) + loga (B ) and loga (AB ) = B · loga (A), for any positive real numbers a, A and B . Problem 4. We have that 2

y = 3x and now we solve for x in terms of y in order to find the inverse function. So, we have (after taking logarithms of both sides) log3 (y) = x2 . Then we take square-roots and obtain x=

p

log3 (y).

So, the inverse function for f (x) is the function g : [1, +∞) −→ [0, +∞) given by the formula p f (x) = log3 (x). Problem 5. For these type of questions, we want to guess the value b = g(a) such that f (b) = f (g(a)) = a. because g(x) is the inverse of f (x). We only guess the value b = g(a) since it is very hard to find the actual formula for g(x), the inverse function of f (x). But guessing the value b = g(a) turns out not to be very difficult. (a) So, we have f (x) = ex + xe and we want to find g (2ee ). So, we want to guess the value b = g (2ee ) such that f (g (2ee )) = 2ee , i.e. f (b) = 2ee . In other words, b satisfies eb + be = 2ee . It’s not hard to guess that b = e works. So, g (2ee ) = e. (b) This time we have f (x) = (1 + x2 )x and we want to find g(2). So, letting b = g(2) then f (g(2)) = 2 and so, f (b) = 2. In other words, b satisfies (1 + b2 )b = 2. It’s not hard to guess that b = 1 works. So, g(2) = 1....