How do you find the equation of a line tangent to the function y=xx at (4,8)?

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Ijezid8t · Accepted Answer

Write the function as:  y=x32  Verify that the point,  (4,8), is on the curve:  8=432  8=8← verified  Compute the first derivative of the function:  dydx=32x12  The slope, m, of the tangent line is the first derivative evaluated at the x coordinate of the point, x=4:  m=32412  m=3  Use the point-slope form of the equation of a line for the tangent line:  y=m(x−x0)+y0  Substitute m=3, x0=4 and y0=8:  y=3(x-4)+8  Simplify:  y=3x-4

Macey Mata · Accepted Answer

f(x)=xxf(x)=x1×12=x32Apply power rule.f′(x)=32x12=3x2∴ the slope of f(x) at x=4 is 3⋅42=±3Since f(x) is defined for x≥0→f′(4)=+3Let y=mx+c be the tangent to f(x) at x=4Since y at x=4 will have the same slope as f(x)→m=3f(4)=4×4=8Thus, 8=3×4+c→c=−4Hence, the equation of our tangent line is: y=3x-4

How do you find the equation of a line tangent

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