Answer:
dfghfegfefehfgfe
Step-by-step explanation:
tyfvrfvewfgwvehgew e
Answer:
A... .................. ....
Answer:
<h2><em><u>2</u></em></h2>
Step-by-step explanation:
<em><u>To</u></em><em><u> </u></em><em><u>find</u></em><em><u> </u></em><em><u>value</u></em><em><u>:</u></em>
2x + 3z
<em><u>Given</u></em><em><u> </u></em><em><u>values</u></em><em><u>:</u></em>
x = 4, y = 3 and z = -2
<em><u>Solution</u></em><em><u>:</u></em>
2x + 3z
<em>(</em><em>Putting</em><em> </em><em>the</em><em> </em><em>values</em><em> </em><em>of</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>4</em><em> </em><em>and</em><em> </em><em>z</em><em> </em><em>=</em><em> </em><em>-2</em><em>)</em>
= 2(4) + 3(-2)
= 8 - 6
= <em><u>2 (Ans)</u></em>
Suppose that some value, c, is a point of a local minimum point.
The theorem states that if a function f is differentiable at a point c of local extremum, then f'(c) = 0.
This implies that the function f is continuous over the given interval. So there must be some value h such that f(c + h) - f(c) >= 0, where h is some infinitesimally small quantity.
As h approaches 0 from the negative side, then:

As h approaches 0 from the positive side, then:

Thus, f'(c) = 0