For all real numbers x, f is a differentiable function such that f(-x) = f(x). Let f(p)=1 and f ' (p) = 5 for some p>0.
1 - Find f ' (-p)
2 - Find f ' (0)
3 - If L1 and L2 are lines tangent to the graph of f at (-p,1) and (p,1), respectively, and is L1 and L2 intersect at point Q, find the x and y coordinates of Q in terms of p.
Note - you don't have to do #3, but it would be much appreciated if you did.
1 - Find f ' (-p)
2 - Find f ' (0)
3 - If L1 and L2 are lines tangent to the graph of f at (-p,1) and (p,1), respectively, and is L1 and L2 intersect at point Q, find the x and y coordinates of Q in terms of p.
Note - you don't have to do #3, but it would be much appreciated if you did.
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f'(-p) = -1
think of a function in which f(-x) = f(x). One example would be x^2, so think of a tangent line in the negative and a tangent line in the positive
f'(0) is 0. Once again, think about the fact that all negative x values of such a fuction will have negative slopes, and all the postive x values will have postive slopes. To go from negative to positive you have to cross 0, the only number that is niether positive or negative is 0, so at 0 slope is 0.
I dont feel like doing number 3, but ill tell you how to do it. Since you know the slope at f(p) and at f(-p) and you have the points that those lines go through, just find the equation of the lines and set them equal to eachother.
think of a function in which f(-x) = f(x). One example would be x^2, so think of a tangent line in the negative and a tangent line in the positive
f'(0) is 0. Once again, think about the fact that all negative x values of such a fuction will have negative slopes, and all the postive x values will have postive slopes. To go from negative to positive you have to cross 0, the only number that is niether positive or negative is 0, so at 0 slope is 0.
I dont feel like doing number 3, but ill tell you how to do it. Since you know the slope at f(p) and at f(-p) and you have the points that those lines go through, just find the equation of the lines and set them equal to eachother.