F. (7) (2 puntos posibles)
Let
f(x)={tanxax+bπ/4<x<π/2x≤π/4
Find a and b such that the function f(x) is differentiable at x=π/4.
To find the values of "a" and "b" such that the function f(x) is differentiable at x = π/4, we need to ensure that the left-hand derivative and the right-hand derivative of f(x) are equal at x = π/4.
The left-hand derivative at x = π/4 is obtained by taking the limit as x approaches π/4 from the left side:
lim (h->0-) [f(π/4 + h) - f(π/4)] / h
We can substitute π/4 into the function and calculate the left-hand derivative:
lim (h->0-) [tan(π/4 + h) - a(π/4) + bπ/4 - tan(π/4)] / h
Since tan(π/4) is equal to 1, we can simplify this expression as:
lim (h->0-) [tan(π/4 + h) - a(π/4) + bπ/4 - 1] / h
= lim (h->0-) [(tan(π/4) + tan(h))/(1 - tan(π/4)tan(h)) - a(π/4) + bπ/4 - 1] / h
Now, we can use the known values of tan(π/4) = 1 and substitute them into the expression:
= lim (h->0-) [(1 + tan(h))/(1 - tan(h))] - a(π/4) + bπ/4 - 1 / h
= lim (h->0-) [(1 + tan(h))/(1 - tan(h))] - (a + 1/4)π + bπ / h
Similarly, we can find the right-hand derivative at x = π/4 by taking the limit as x approaches π/4 from the right side:
lim (h->0+) [f(π/4 + h) - f(π/4)] / h
= lim (h->0+) [tan(π/4 + h) - a(π/4) + bπ/4 - tan(π/4)] / h
Using the same steps as before, we simplify this expression as:
= lim (h->0+) [(tan(π/4) + tan(h))/(1 - tan(π/4)tan(h))] - a(π/4) + bπ/4 - 1 / h
= lim (h->0+) [(1 + tan(h))/(1 - tan(h))] - (a + 1/4)π + bπ / h
Now, in order for f(x) to be differentiable at x = π/4, the left-hand derivative and the right-hand derivative must be equal. Therefore, we can set the two expressions equal to each other:
lim (h->0-) [(1 + tan(h))/(1 - tan(h))] - (a + 1/4)π + bπ / h = lim (h->0+) [(1 + tan(h))/(1 - tan(h))] - (a + 1/4)π + bπ / h
By comparing the constants and coefficients on both sides of the equation, we can find the values of "a" and "b" such that the equation holds true.