find a point on the curve y=x3-11x+5at which the tangent is y= x-11.
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The gradient of y = x³ - 11x + 5
dy/dx = 3x² - 11
The gradient of y = x - 11 is 1
This is a linear equation in the form y = mx + c where m is the gradient
∴ dy/dx = 1 (The gradient of the curve equals gradient of line at point of contact)
3x² - 11 = 1
3x² = 12
x² = 4
x = 2 or -2
if x = 2
y = x³ - 11x + 5
y = 2³ - 11(2) + 5 = -9
∴ The point is (2, -9)
if x = -2
y = (-2)³ -11(-2) + 5 = 19
and the point becomes (-2, 19)
Thank you
dy/dx = 3x² - 11
The gradient of y = x - 11 is 1
This is a linear equation in the form y = mx + c where m is the gradient
∴ dy/dx = 1 (The gradient of the curve equals gradient of line at point of contact)
3x² - 11 = 1
3x² = 12
x² = 4
x = 2 or -2
if x = 2
y = x³ - 11x + 5
y = 2³ - 11(2) + 5 = -9
∴ The point is (2, -9)
if x = -2
y = (-2)³ -11(-2) + 5 = 19
and the point becomes (-2, 19)
Thank you
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