What is the equation of the line normal to #f(x)= sqrt(2x^3-x) # at #x=1#?
The tangent line is of the form:
So the equation of the tangent is:
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To find the equation of the line normal to f(x) at x=1, we need to find the derivative of f(x) and evaluate it at x=1. Then, we can determine the slope of the normal line.
The derivative of f(x) can be found using the power rule and chain rule. Taking the derivative of f(x) = sqrt(2x^3 - x), we get:
f'(x) = (3x^2 - 1) / (2 * sqrt(2x^3 - x))
Evaluating f'(x) at x=1, we have:
f'(1) = (3(1)^2 - 1) / (2 * sqrt(2(1)^3 - 1)) = (3 - 1) / (2 * sqrt(2 - 1)) = 2 / (2 * 1) = 1
The slope of the normal line is the negative reciprocal of the derivative at x=1. Therefore, the slope of the normal line is -1.
Now, we have the slope (-1) and a point on the line (x=1, f(1)). We can use the point-slope form of a line to find the equation of the line normal to f(x) at x=1.
Using the point-slope form, the equation of the line is:
y - f(1) = -1(x - 1)
Simplifying further, we get:
y - f(1) = -x + 1
This can be rearranged to obtain the equation of the line normal to f(x) at x=1:
y = -x + f(1) + 1
Therefore, the equation of the line normal to f(x) = sqrt(2x^3 - x) at x=1 is y = -x + f(1) + 1.
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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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