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For f(x)=1/5+x^2, the slope of the graph of y=f(x) is known to be -1/18 at the point with x-coordinate 1. Find the equation of the tangent line at the point.

User Simcha
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Explanation:

To find the equation of the tangent line at the point with x-coordinate 1, we need to determine both the slope and the y-coordinate of the point on the graph of the function f(x) = 1/5 + x^2.

Given that the slope of the graph at the point (1, f(1)) is -1/18, we can differentiate the function f(x) with respect to x to find the derivative.

f'(x) = d/dx (1/5 + x^2)

= 2x

Next, we can evaluate the derivative at x = 1 to find the slope at that point:

f'(1) = 2(1)

= 2

Since the slope of the graph at the point (1, f(1)) is -1/18, we can set up the following equation:

2 = -1/18

Solving for the y-coordinate:

f(1) = 1/5 + 1^2

= 1/5 + 1

= 6/5

Therefore, the y-coordinate of the point on the graph is 6/5.

Now that we have both the slope (-1/18) and the y-coordinate (6/5) of the point (1, f(1)), we can use the point-slope form of the equation of a line to find the equation of the tangent line.

y - y1 = m(x - x1)

Substituting the values:

y - (6/5) = (-1/18)(x - 1)

Simplifying:

y - 6/5 = (-1/18)x + 1/18

Bringing y to the left side:

y = (-1/18)x + 1/18 + 6/5

y = (-1/18)x + (1/18 + 36/18)

y = (-1/18)x + 37/18

Therefore, the equation of the tangent line to the graph of f(x) = 1/5 + x^2 at the point (1, 6/5) is y = (-1/18)x + 37/18.

User Captaintom
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