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Sergeu [11.5K]
3 years ago
5

For what value of the number k is the following function differentiable at x=2

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0

Answer:

For f(x) to be differentiable at 2, k = 5.

Step-by-step explanation:

For f(x) to be differentiable at x = 2, f(x) has to be continuous at 2.

For f(x) to be continuous at 2, the limit of f(2 – h) = f(2) = f(2 + h) as h tends to 0.

Now,

f(2 – h) = 2(2 – h) + 1 = 4 – 2h + 1 = 5 – 2h.

As h tends to 0, lim (5 – 2h) = 5

Also

f(2 + h) = 3(2 + h) – 1 = 6 + 3h – 1 = 5 + 3h

As h tends to 0, lim (5 + 3h) = 5.

So, for f(2) to be continuous k = 5

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Read 2 more answers
Please help!! thank you.
Zina [86]

(A) Product of -2x^{3}+x-5 and x^{3}-3x-4 is:

(-2x^{3}+x-5)(x^{3}-3x-4)

=-2x^{3}(x^{3}-3x-4)+x(x^{3}-3x-4)-5(x^{3}-3x-4)

=(-2x^{3})(x^{3})+(-2x^{3})(-3x)+(-2x^{3})(-4)+x(x^{3})+(x)(-3x)+(x)(-4)+(-5)(x^{3})+(-5)(-3x)+(-5)(-4)

=-2x^{6}+6x^{4}+8x^{3}+x^{4}-3x^{2}-4x-5x^{3}+15x+20

=-2x^{6}+6x^{4}+x^{4}+ 8x^{3}-5x^{3}-3x^{2}-4x+15x+20

=-2x^{6}+7x^{4}+3x^{3}-3x^{2}+11x+20

(B) Yes.

Product of -2x^{3}+x-5 and x^{3}-3x-4 = Product of x^{3}-3x-4 and -2x^{3}+x-5 because multiplication is commutative.

Commutative Property of multiplication says that a.b = b.a.

Thus, multiplication is same irrespective of the order of two numbers.



5 0
3 years ago
Write an algebraic expression to represent each verbal expression. 5 times the sum of a number and 1
11111nata11111 [884]

Answer: 5(x+1)

Step-by-step explanation:

Let the number be x

Sum of number and 1 is x+1

5 times the sum is 5(x+1)

3 0
3 years ago
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