Question

数学题有没有会的

Answer 1

问：数学题有没有会的

答：i) To find the gradient of the function at x = 2, we need to differentiate the function with respect to x. So,y = x^3 - 4x - 9ady/dx = 3x^2 - 4At x = 2, dy/dx = 3(2)^2 - 4 = 8.So, the gradient of the function at x = 2 is 8.ii) To find the equation of the tangent at x = 2, we use the point-slope form of the equation of a line.y - y1 = m(x - x1)where m is the gradient of the tangent and (x1, y1) is the point of tangency.We know that the point of tangency is (2, y(2)), where y(2) is the value of the function at x = 2. So,y(2) = 2^3 - 4(2) - 9a = -9a + 8And, we know that the gradient of the tangent at x = 2 is 8 (from part i). So,y - (-9a + 8) = 8(x - 2)y = 8x - 25aTherefore, the equation of the tangent at x = 2 is y = 8x - 25a.b) To find the stationary point of the curve y = 3x^2 - 12x + 1, we need to find where the gradient of the function is zero. So,y = 3x^2 - 12x + 1dy/dx = 6x - 12Setting dy/dx = 0, we get6x - 12 = 0Solving for x, we get x = 2.S

答：a) To find the gradient of the function at x = 3, we need to differentiate the function with respect to x. So,y = x^2 - 5x + 2dy/dx = 2x - 5At x = 3, dy/dx = 2(3) - 5 = 1.So, the gradient of the function at x = 3 is 1.b) The normal line is perpendicular to the tangent line at the point of tangency. Therefore, the gradient of the normal line is the negative reciprocal of the gradient of the tangent line.The gradient of the tangent line at x = 3 is 1 (from part a), so the gradient of the normal line at x = 3 is -1/1 = -1.c) To find the equation of the normal line at x = 3, we use the point-slope form of the equation of a line.y - y1 = m(x - x1)where m is the gradient of the normal line and (x1, y1) is the point of tangency.We know that the point of tangency is (3, y(3)), where y(3) is the value of the function at x = 3. So,y(3) = 3^2 - 5(3) + 2 = -4And, we know that the gradient of the normal line at x = 3 is -1 (from part b). So,y - (-4) = -1(x - 3)y = -x + 1