Please help thank you

Mathematics

1 answer:

Keith_Richards [23]3 years ago

7 0

Answer:

alternates

Step-by-step explanation:

The series is for k = 0 to 4, so k = 0, 1, 2, 3, 4.

That is 5 terms.

-1/2 is inside the parentheses and is raised to 0, 1, 2, 3, 4.

A negative number raised to an even integer is positive.

A negative number raised to an odd integer is negative.

The sign of each term in the series alternates.

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<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$