Help on differentiable functions

If a is such that f(x) is continuous at x=1, is f(x) also differentiable at x=1? Justify your answer.

Let f(x)= {x+2a if x<1
{ax^2+7x-4 if x≥1

If f(x) is continuous then the following must be true:

lim[x-->1] f(x) = f(1)

for lim[x-->1] f(x) to exist, lim[x-->1 (-)] f(x) = lim[x-->1 (+)] f(x) ----- I will use (-) or (+) to signify the limits from left or from right.

lim[x-->1 (-)] f(x) = lim[x-->1 (+)] f(x)
lim[x-->1 (-)] (x + 2a) = lim[x-->1 (+)] (ax^2 + 7x - 4)
1 + 2a = a + 7 - 4
a = 2

so f(x) = {x + 4 if x<1
{2x^2 + 7x - 4 if x>=1

lim[x-->1] f(x) = 5
f(1) = 5
therefore lim[x-->1] f(x) = f(1), so the function is continuous only when a = 2.

the derivative at x = 1 is:

lim[h-->0] [(f(1 + h) - f(1)) / h]
for this limit to exist, the limit from both sides must be equal.
the limit from the left is:

lim[h-->0 (-)] [(f(1 + h) - f(1)) / h]
= lim[h-->0 (-)] [((1+ h + 4) - 5) / h]
= lim[h-->0 (-)] [h / h]
= 1

the limit from the right is:

lim[h-->0 (+)] [(f(1 + h) - f(1)) / h]
= lim[h-->0 (+)] [((2(1+ h)^2 + 7(1 + h) - 4) - 5) / h]
= lim[h-->0 (-)] [(2 + 4h + 2h^2 + 7 + 7h - 4 - 5) / h]
= lim[h-->0 (-)] [(2h^2 + 11h) / h]
= lim[h-->0 (-)] [2h + 11]
= 11

therfore, since the limit from both sides are not equal, the limit does not exist. Which means the derivative at x = 1 does not exist. The function is not differentiable at x = 1.

f(1) = a + 7 - 4 = a + 3
limit[x→1⁻] f(x) = lim[x→1⁻] (x + 2a) = 1 + 2a
limit[x→1⁺] f(x) = lim[x→1⁺] (ax²+7x-4) = a + 3

Since f(x) is continuous, then lim[x→1⁻] = lim[x→1⁺] = f(1)
1 + 2a = a + 3
a = 2

f(x) =
{ x + 4 . . . . . . . if x < 1
{ 2x² + 7x - 4 . . if x ≥ 1

-------------------------------

f'(x) =
{ 1 . . . . . . if x < 1
{ 4x + 7 . . if x ≥ 1

limit[x→1⁻] f'(x) = lim[x→1⁻] (1) = 1
limit[x→1⁺] f'(x) = lim[x→1⁺] (4x + 7) = 11

lim[x→1⁻] ≠ lim[x→1⁺] -----> f(x) is not differentiable at x = 1


Mαthmφm