What is the equation of the line that is normal to #f(x)= x-sqrt( 3x+2) # at # x=4 #?
The tangent and normal are orthogonal to each other Slope of the tangent is 0.599 m1m2=-1gives Slope of the normal is -1/0.599=-1.669 The normal passes through the point (4,0.258) Equation of the line passing through a point having a slope is given by
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To find the equation of the line that is normal to the function f(x) = x - sqrt(3x + 2) at x = 4, we need to determine the slope of the tangent line at x = 4 and then find the negative reciprocal of that slope to obtain the slope of the normal line.
First, we find the derivative of f(x) with respect to x:
f'(x) = 1 - (3/2)(3x + 2)^(-1/2)
Next, we substitute x = 4 into f'(x) to find the slope of the tangent line at x = 4:
f'(4) = 1 - (3/2)(3(4) + 2)^(-1/2)
Simplifying this expression, we get:
f'(4) = 1 - (3/2)(14)^(-1/2)
Now, we can find the negative reciprocal of f'(4) to obtain the slope of the normal line:
Slope of normal line = -1 / f'(4)
Finally, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of tangency (in this case, (4, f(4))), and m is the slope of the normal line, to find the equation of the line that is normal to f(x) at x = 4.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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