Sept 20, 2011 MATH 140: Calculus I Tutorial 2. ln(x 2 1) = 3 x 2 1 = e 3 x = e 3 + 1

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1 Sept, MATH 4: Calculus I Tutorial Solving Quadratics, Dividing Polynomials Problem Solve for x: ln(x ) =. ln(x ) = x = e x = e + Problem Solve for x: e x e x + =. Let y = e x. Then we have a quadratic equation, y y + =. By solving the quadratic equation, we find that the roots are y =,. In other words, we must solve e x = and e x =. But that means that x = {, ln()}. Problem Find the domain and range of the following function: f(x) = 4 x. We know that 4 x has to be positive in order to avoid imaginary numbers when taking the square root. This happens for the domain x [, ]. To determine the range, note that when x =,, f(x) =. Clearly f(x) reaches its maximal value when x = ; here f(x) =. So the range is [, ]. Problem 4 Find the domain and range of the following function: h(t) = 4 t t. Let s first simplify: h(t) = 4 t t = ( t)( + t) t = + t There are no restrictions to the domain of this simplified function. The domain of that function is the whole real line. However, consider the non-simplified function at t =. Here h() = this is undefined. So the domain of h(t) is R \ (the whole real line except for the point ). This missing point will affect the range. At t =, the simplified function has value + = 4. So the range is R \ {4} (or, written differently, (, 4) (4, )). Page of 7

2 Sept, MATH 4: Calculus I Tutorial Problem 5 Complete the square: x + x. Let a be a real number. Then (x + a) = x + ax + a. In our problem, a =, or a =. So x + x = (x + ) 9 4. Problem 6 Complete the square: x + x = 6, so let s first expand (x + 6) : x + x + 6. In our case, x + x + 6 = (x + 6) 9x. Problem 7 Complete the square: 4x + 9x x + 9x + 6 = 4(x x + 4). This is close to 4(x + ), so let s expand this expression and see what we need to take away. 4(x + ) = 4(x + 4x + 4) = 4x + 6x + 6 For this problem, we find 4x + 9x + 6 = 4(x + ) 7x. Problem 8 Complete the square: 5x + x x + x + 5 = 5(x + 5 x + ). But 5(x + ) = 5x + x + 5. In this case, 5x + x + 5 = 5(x + ) + ( )x. Problem 9 Solve the quadratic equation x 4x 6 =. Using the quadratic formula, 4± 6 4( 6) = ± 5. Page of 7

3 Sept, MATH 4: Calculus I Tutorial Problem Solve the quadratic equation x + x 4 =. Using the quadratic formula, ± 9 4( 4) = 4,. Problem Find the roots of the following cubic x 7x 6 =, where one root is given to be. x + x +. x ) x 7x 6 x + x x 7x x + 9x Problem Find x +x x+. x 6 x + 6 x x +. x + ) x + x x x x x + x + 9x x x 6 46 So the solution is x x + 46 x+. Page of 7

4 Sept, MATH 4: Calculus I Tutorial Solving Inequalities Problem Find where this inequality holds: x 4 < (7 x)(x ). The right hand side of this equation is only valid where (7 x)(x ). So let s find the range of x that satisfies this inequality first. Clearly, the equation (7 x)(x ) = at x = 7,. So we can divide the real line into three chunks: (, ), [, 7], and (7, ). You can easily make a table or plug in points from each of these domains to see that (7 x)(x ) when x [, 7]. Now we can search for the solution to the following inequality on [, 7]: (x 4) < (7 x)(x ) However, we only care about when x 4 now. Why? Because x 4 < < (7 x)(x ) when x < 4 (The left hand side is negative, and the right hand side is positive, positive numbers are always greater than negative numbers, so the inequality is true!). We know that the range [, 4) makes this inequality true. We just want to find out where the inequality holds for x (4, 7]. Let s find where the right hand side and left hand side are equal. To do this, we square both sides and add the restriction that x 4. Luckily, we already have this restriction! So we compute: (x 4) = (7 x)(x ) x 8x + 6 = x + 9x 4 x 7x + = Using the quadratic formula, the roots of this equation are at x =.5, 6. But we are only looking for intersections in the interval [4, 7], so we only need to consider the point x = 6. By plugging in points (or drawing the graph!), we see that the equality holds for x [4, 6), and does not hold above 6. Combining this interval and the interval we found previously, we know the solution is x [, 6). Plot showing the inequality is true for [,6) 4 x-4 ((7-x)(x-)) / f(x) x Page 4 of 7

5 Sept, MATH 4: Calculus I Tutorial Problem 4 This is Problem from the sample quiz. Find for which x the following inequality is true: e x 5e x <. We know that if a < b, then b < a < b. Let s expand our absolute value similarly: < e x 5e x < Now let s make a change of variables, y = e x, to make this a polynomial equation. With this change of variables, we must be careful to specify that y >, as the exponential function (e x ) is never negative! Now we have the following: < y(y 5) < 5 The equation y(y 5) has roots at y =,. You can see this in the figures below as well. Between those roots, the function is negative, and it s positive elsewhere. In our case, there are two places that the function and the line f(y) = intersect, and two places that the function and the line f(y) = intersect. We can set y(y 5) equal to these values to find the points of intersection. 5 4 Graph of the inequality y -5y - Another way to visualize the inequality y -5y f(y) f(y) y y First let s deal with y(y 5) =. Simplifying this, we find y 5y =, a quadratic equation. You may factor this, or you can use the quadratic formula to solve for y: By simplifying this, one finds that y =,. y = 5 ± 5 4()( ) Similarly, set y(y 5) =. By simplifying we find y 5y + =. Once again using the quadratic equation, we find that y =,. Therefore we conclude that this inequality is valid for y (, ) (, ). However we must add the original constraint that y must be positive. Therefore y (, ) (, ). Now we can transform back to x, since if y = e x, then x = ln(y). The solution is that the inequality is valid for x (, ln( )) (, ln()). Convince yourself that ln( ) = ln( ). Page 5 of 7

6 Sept, MATH 4: Calculus I Tutorial Trig Functions Problem 5 Convert from radians to degrees: 4π. 4π( 8 π ) = 7. Problem 6 Find the other ratios (cos θ, tan θ): sin θ = 5, < θ < π/. Since sin θ = opposite hypotenuse, we know that the side of the triangle opposite θ has length, and that the hypotenuse has length 5. Call the missing side a (the side adjacent to θ). Then by the Pythagorean Theorem, a + = 5, so a = 4. Therefore cos θ = 4 5 and tan θ = 4. Problem 7 Prove sin θ cot θ = cos θ. sin θ cot θ = sin θ tan θ = sin θ cos θ sin θ = cos θ. Problem 8 Prove sin x sin x + cos x cos x = cos x. Using an identity and that cos(x) is even: sin x sin x + cos x cos x = [cos(x x) cos(x)] + [cos(x) + cos(x x)] = cos( x) = cos(x). Page 6 of 7

7 Sept, MATH 4: Calculus I Tutorial Problem 9 Solve for x: sin x =, x [, π]. Taking the to the right hand side and applying the square root we simplify to: sin x = ± Solution is at the intersection..8 sin(x) (/) / -(/) /.6.4 f(x) x Clearly the solution is x = { π 6, π π 6, π + π 6, π π 6 } = { π 6, 5π 6, 7π 6, π 6 }. Final Note: To discuss Problem from the homework, it s best to come to office hours. Page 7 of 7