a tangent to the graph of $y = x^{2}+1$ is parallel to the line $y = x$. find the $y$-coordinate of the…

Explanation:

Step1: Find the derivative of the function

The function is (y = x^{2}-\frac{1}{x}). Using the power rule ((x^{n})^\prime=nx^{n - 1}) and ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}) (here (u = 1), (v=x)), the derivative (y^\prime=2x+\frac{1}{x^{2}}).

Step2: Set the derivative equal to the slope of the given line

The line (y = x) has a slope (m = 1). So we set (2x+\frac{1}{x^{2}}=1). Multiply through by (x^{2}) (assuming (x\neq0)) to get (2x^{3}+1=x^{2}), or (2x^{3}-x^{2}+1 = 0). We can try some simple integer values. When (x=-1), (2(-1)^{3}-(-1)^{2}+1=2\times(-1)-1 + 1=-2\neq0). When (x = 1), (2(1)^{3}-(1)^{2}+1=2 - 1+1=2\neq0). Let's rewrite the equation as (2x^{3}-x^{2}+1=(x + 1)(2x^{2}-3x + 1)). Factoring (2x^{2}-3x + 1=(2x - 1)(x - 1)). The roots of (2x^{3}-x^{2}+1 = 0) are (x=-1,\frac{1}{2},1). Substitute (x=\frac{1}{2}) into (y = x^{2}-\frac{1}{x}), (y=(\frac{1}{2})^{2}-\frac{1}{\frac{1}{2}}=\frac{1}{4}-2=-\frac{7}{4}) (not an integer). Substitute (x = 1) into (y = x^{2}-\frac{1}{x}), (y=1^{2}-\frac{1}{1}=0) (an integer).

Answer:

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