The first derivative of the function f(x) = x² - 5 is equal to f'(x) = 2 · x.
<h3>How to find the derivative of a quadratic equation by definition of derivative</h3>
In this question we have a quadratic function, in which we must make use of the definition of derivative to find the expression of its first derivative. Then, the procedure is shown below:
f(x) = x² - 5 Given
f' = [(x + h)² - 5 - x² + 5] / h Definition of derivative
(x² + 2 · x · h + h² - 5 - x² + 5) / h Perfect square trinomial
(2 · x · h + h²) / h Associative, commutative and modulative properties / Existence of additive inverse
2 · x + h Distributive, commutative and associative properties / Definition of division / Existence of multiplicative inverse
2 · x h = 0 / Result
The first derivative of the function f(x) = x² - 5 is equal to f'(x) = 2 · x.
To learn more on derivatives: brainly.com/question/25324584
#SPJ1
<h3>
Answer: 0.6</h3>
========================================
Work Shown:
sin(angle) = opposite/hypotenuse
sin(T) = VU/VT
sin(T) = 3/5
sin(T) = 0.6
Answer:
5 2/5
Step-by-step explanation:
(2 1/4) / x = (1 1/2) / (3 3/5)
so
x = (2 1/4) * (3 3/5) / (1 1/2)
Answer:
The value of c = -0.5∈ (-1,0)
Step-by-step explanation:
<u>Step(i)</u>:-
Given function f(x) = 4x² +4x -3 on the interval [-1 ,0]
<u> Mean Value theorem</u>
Let 'f' be continuous on [a ,b] and differentiable on (a ,b). The there exists a Point 'c' in (a ,b) such that
<u>Step(ii):</u>-
Given f(x) = 4x² +4x -3 …(i)
Differentiating equation (i) with respective to 'x'
f¹(x) = 4(2x) +4(1) = 8x+4
<u>Step(iii)</u>:-
By using mean value theorem
8c+4 = -3-(-3)
8c+4 = 0
8c = -4
c ∈ (-1,0)
<u>Conclusion</u>:-
The value of c = -0.5∈ (-1,0)
<u></u>
Answer:
203 Ibs of butter
Step-by-step explanation:
