Dhwani Chovatia Assignment WW4 due 10/06/2023 at 11:59pm PDT Problem 1. (1 point) () If f(x) = 4x8 - 4x5 - 2x3 +4x, find f'(x) and f'(1). f'(x) = f'(1) = Correct Answers: . 32x7 - 20x4 - 6x2 + 4 . 10 Problem 2. (1 point) () Let f(x) = 3x2-4x+2. Compute f'(-2). Answer: Use this to find the equation of the tangent line to the parabola y = 3x2 - 4x +2 at the point (-2, 22). The equation of this tangent line can be written in the form y = mx +b. Determine m and b. m = b = Correct Answers: · - 16 · - 16 · - 10 Problem 3. (1 point) () Differentiate: Y(u)=(u-2+u-3) (u5 - 3u2) Y'(u) = Correct Answers: · (-2u-3 -31-4) (45 +-3u2) +(u-2 +1-3) (Su4 +2. (-3) u) 1 2023W1 MATH 100C ALL 2023W1 Problem 4. (1 point) () Differentiate: F(v)=(1-4) (x + 10.3) F'(y) = Correct Answers: . 10+. -1.10-1 + 3. (-1) y2 y 4 Problem 5. (1 point) () Let f(x) = 7x3(x2 -7). Evaluate f'(x) at the following points: (A) f'(11) = (B) f'(-1)= Correct Answers: . 3.7 . 112 (112 -7) +7 . 113 . 2 . 11 . 3.7. (-1)2 ((-1)2-7)+7.(-1)3.2.(-1) Problem 6. (1 point) () Find the equation of the tangent line to the curve y = x/x at the point (9,27). y = Solution: Rather than treating xx as the product of x and Vx, and using the product rule, it's easier to simplify it to x3/2. (Both methods give the same result, but the second one is faster.) f(x)=xVx = x3/2 f(x)= 2x1/2 = 2Vx f'(9) = 2V9 This gives us the slope of our tangent line. The point it passes through is (9,27). So, starting in point-slope form, our tangent line is: y - 27 = 25 (x-9) which is equivalent to 3 y=27+-v9(x-9). Correct Answers: • 2 2 .97(x-9)+27 1
Problem 7. (1 point) () Find the parabola with equation y = ax2 + bx whose tangent line at (1,4) has equation y = 6x - 2. a = b = Solution: There are two unknown parameters to find (a and b), and the question gives us two pieces of information: a point the function passes through, and its tangent line at that point. This will lead to a system of two equations with two variables. 1. Point To even have a tangent line to the curve at the point (1,4), the curve must pass through that point. So, this gives us our first equa- tion: y(1) = 4. That is, a + b = 4 . 2. Slope The tangent line has equation 6x - 2, so it has slope 6. From y = ax2 + bx, we see y'(x) = 2ax +b, and so y'(1) = 2a+b. So, the tangent line to the curve at x = 1 has slope 2a +b. This gives us our second equation: 2a +b = 6. Solution There are various