Josh D.
asked • 02/27/24Find the equations of all lines that are tangent to the graph of the function f(x) = x ^ 2 + 1 and pass through the point P(1,0). Be careful: this point is not on the graph of f.
You want the point (x,y) to be such that for (x,x2+1) id s y a tangent with a slope of f' = 2x (same slope as line from (1,0) to (x,y))
The slope is 2x = (x2+1 - 0)/(x-1) or x = 1 ± sqrt(2)
Eqns of lines are 2(1±sqrt(2)) = y/(x-1) (This can be rearranged to ther forms)
Doug C. answered • 02/27/24
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Josh D.
I have this question I got the answer but I think I am wrong cause idk what it means by [0, infinity). If f(x) = (e ^ x + e ^ (- x))/2 with domain confined to the interval [0, infinity), find the derivative of the inverse function by using the formula (f ^ - 1)' * (x) = 1/(f' * (f ^ - 1 * (x)))02/27/24
Doug C.
Take a look at this graph. Without the restriction on the domain the function does not have an inverse FUNCTION because it is not 1-1 (does not pass the horizontal line test). The restriction [0, inf) is a way to insure that an inverse function exists. desmos.com/calculator/cp56xnux3z02/27/24
Doug C.
By the way, in order to do this problem you have to know the inverse function which is ln(x + sqrt(x^2-1)).02/28/24
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Josh D.
thank you so much!!! it makes so much sense with the graph. I was super confused on this question but now it makes great sense!. thank you again!!02/27/24