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Given that the image of the function f (x) = AX2 + BX3 passes through point m (1,4), the tangent of the curve at point m is just perpendicular to the straight line x + 9y = 0 1. Find the value of real numbers a and B 2. Calculate the value range of the slope of the tangent at a certain point on the known function image
Substituting (1,4), a + B = 4
The derivative of F (x) is 2aX + 3bx & # 178; when x = 1, the derivative is - 1 / (- 1 / 9) = 9
Then 2A + 3B = 9, ②,
From the solution of (1) and (2), a = 3, B = 1
The range of tangent slope is 2aX + 3bx & # 178; the range is a, B, and the range is greater than or equal to - 3
Given that the image of the function f (x) = ax ^ 3 = BX ^ 2 passes through the point m (1,4), the tangent of the curve at the point m is just perpendicular to the straight line x = 9y = 0. (1) find the value of the real number A.B
If the individual equal sign of your question above is a plus sign
Then. A = 1, B = 3
A + B = 4... 3A + 2B = 9
So get the answer
Given that the function f (x) = ax to the third power plus BX passes through the point m (1,4), the tangent at the point m is just perpendicular to the straight line x + 9y + 5 = 0. Find the value of a and B
Because f (x) = ax & sup3; + BX & sup2; passes through M (1,4) ∵ 4 = a + B (1) and ∵ its tangent at point m is just perpendicular to the straight line x + 9y + 5 = 0, and the slope of the straight line x + 9y + 5 = 0 is - 1 / 9
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