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3 years ago

5

Let the function f be continuous and differentiable for all x. Suppose you are given that f(−1)=−3, and that f'(x) for all value

s of x. Use the Mean Value Theorem to determine the largest possible value of f(5).

1 answer:

Monica [59]3 years ago

8 0

Answer:

Step-by-step explanation:

By MVT,

$f'(c) = \frac{f(5)-f(-1)}{5-1}$

$4f'(c)= f(5) + 3\rightarrow f(5) = 4f'(c) - 3$

Moreover, $f'(x)\leq 4~~~~~~ \forall x,$

$f(5) \leq 16 - 3 =13$

Hence the largest possible value of f(5) is 13

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Find a set of parametric equations for y= 5x + 11, given the parameter t= 2 – x

Sati [7]

Answer:

$x = 2-t$ and $y = -5\cdot t +21$

Step-by-step explanation:

Given that $y = 5\cdot x + 11$ and $t = 2-x$ , the parametric equations are obtained by algebraic means:

1) $t = 2-x$ Given

2) $y = 5\cdot x +11$ Given

3) $y = 5\cdot (x\cdot 1)+11$ Associative and modulative properties

4) $y = 5\cdot \left[(-1)^{-1} \cdot (-1)\right]\cdot x +11$ Existence of multiplicative inverse/Commutative property

5) $y = [5\cdot (-1)^{-1}]\cdot [(-1)\cdot x]+11$ Associative property

6) $y = -5\cdot (-x)+11$ $\frac{a}{-b} = -\frac{a}{b}$ / $(-1)\cdot a = -a$

7) $y = -5\cdot (-x+0)+11$ Modulative property

8) $y = -5\cdot [-x + 2 + (-2)]+11$ Existence of additive inverse

9) $y = -5 \cdot [(2-x)+(-2)]+11$ Associative and commutative properties

10) $y = (-5)\cdot (2-x) + (-5)\cdot (-2) +11$ Distributive property

11) $y = (-5)\cdot (2-x) +21$ $(-a)\cdot (-b) = a\cdot b$

12) $y = (-5)\cdot t +21$ By 1)

13) $y = -5\cdot t +21$ $(-a)\cdot b = -a \cdot b$ /Result

14) $t+x = (2-x)+x$ Compatibility with addition

15) $t +(-t) +x = (2-x)+x +(-t)$ Compatibility with addition

16) $[t+(-t)]+x= 2 + [x+(-x)]+(-t)$ Associative property

17) $0+x = (2 + 0) +(-t)$ Associative property

18) $x = 2-t$ Associative and commutative properties/Definition of subtraction/Result

In consequence, the right answer is $x = 2-t$ and $y = -5\cdot t +21$ .

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Answer:

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Step-by-step explanation:

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Required

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