# What is the equation of the normal line of #f(x)=x^3-x^2+17x# at #x=7#?

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is

so If

# \ \ \ \ \ f'(x) = 3x^2-2x+17 #

When

So the tangent passes through

# y-413= -1/150(x-7) #

# \ \ \ \ \ :. y = -1/150x +61957/150 #

We can confirm this solution is correct graphically:

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To find the equation of the normal line at a given point on a curve, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope.

To find the slope of the tangent line, we take the derivative of the function f(x).

The derivative of f(x) = x^3 - x^2 + 17x is f'(x) = 3x^2 - 2x + 17.

To find the slope of the tangent line at x = 7, we substitute x = 7 into the derivative: f'(7) = 3(7)^2 - 2(7) + 17 = 147 - 14 + 17 = 150.

The slope of the tangent line at x = 7 is 150.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line: -1/150.

Now, we have the slope of the normal line and the point (7, f(7)) = (7, 7^3 - 7^2 + 17(7)) = (7, 343 - 49 + 119) = (7, 413).

Using the point-slope form of a line, the equation of the normal line is y - 413 = (-1/150)(x - 7).

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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.

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