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3 years ago

Solve the system by graphing. It is REQUIRED to check your solution. Show work for this problem on your work page.

Mathematics

2 answers:

Hunter-Best [27]3 years ago

6 0

Answer:

y=x-1

y=-2x-4

although I cant summon a graph for this one, I can give cords

for first graph (-2,-3),(-1,-2),(0,-1), (1,0),(2,1)

For second graph the slope is down 2 over 1, and begins at (0,-4).

(-2,0)(-1,-2),(0,-4),(1,-6),(2,-8)

Rashid [163]3 years ago

5 0

Let's solve for x.

y=x−1

Step 1: Flip the equation.

x−1=y

Step 2: Add 1 to both sides.

x−1+1=y+1

x=y+1

Answer:

x=y+1

---------------------------------

Let's solve for x.

y=−2x−4

Step 1: Flip the equation.

−2x−4=y

Step 2: Add 4 to both sides.

−2x−4+4=y+4

−2x=y+4

Step 3: Divide both sides by -2.

−2x

−2

y+4

−2

−1

y−2

Answer:

−1

y−2

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Need help on both of these Calculus AB mock exams.

babymother [125]

AB1

(a) $g$ is continuous at $x=3$ if the two-sided limit exists. By definition of $g$ , we have $g(3)=f(3^2-5)=f(4)$ . We need to have $f(4)=1$ , since continuity means

$\displaystyle\lim_{x\to3^-}g(x)=\lim_{x\to3}f(x^2-5)=f(4)$

$\displaystyle\lim_{x\to3^+}g(x)=\lim_{x\to3}5-2x=-1$

By the fundamental theorem of calculus (FTC), we have

$f(6)=f(4)+\displaystyle\int_4^6f'(x)\,\mathrm dx$

We know $f(6)=-4$ , and the graph tells us the *signed* area under the curve from 4 to 6 is -3. So we have

$-4=f(4)-3\implies f(4)=-1$

and hence $g$ is continuous at $x=3$ .

(b) Judging by the graph of $f'$ , we know $f$ has local extrema when $x=0,4,6$ . In particular, there is a local maximum when $x=4$ , and from part (a) we know $f(4)=-1$ .

For $x>6$ , we see that $f'\ge0$ , meaning $f$ is strictly non-decreasing on (6, 10). By the FTC, we find

$f(10)=f(6)+\displaystyle\int_6^{10}f'(x)\,\mathrm dx=7$

which is larger than -1, so $f$ attains an absolute maximum at the point (10, 7).

$k(3)=\dfrac{\displaystyle3\int_4^9f'(t)\,\mathrm dt-12}{3e^{2f(3)+5}-3}$

From the graph of $f'$ , we know the value of the integral is -3 + 7 = 4, so the numerator reduces to 0. In order to apply L'Hopital's rule, we need to have the denominator also reduce to 0. This happens if

$3e^{2f(3)+5}-3=0\implies e^{2f(3)+5}=1\implies 2f(3)+5=\ln1=0\implies f(3)=-\dfrac52$

Next, applying L'Hopital's rule to the limit gives

$\displaystyle\lim_{x\to3}k(x)=\lim_{x\to3}\frac{9f'(3x)-4}{6f'(x)e^{2f(x)+5}-1}=\frac{9f'(9)-4}{6f'(3)e^{2f(3)+5}-1}=-\dfrac4{21e^{2f(3)+5}-1}$

which follows from the fact that $f'(9)=0$ and $f'(3)=3.5$ .

Then using the value of $f(3)$ we found earlier, we end up with

$\displaystyle\lim_{x\to3}k(x)=-\dfrac4{20}=-\dfrac15$

(d) Differentiate both sides of the given differential equation to get

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=2f''(2x)(3y-1)+f'(2x)\left(3\dfrac{\mathrm dy}{\mathrm dx}\right)=[2f''(2x)+3f'(2x)^2](3y-1)$

The concavity of the graph of $y=h(x)$ at the point (1, 2) is determined by the sign of the second derivative. For $y>\frac13$ , we have $3y-1>0$ , so the sign is entirely determined by $2f''(2x)+3f'(2x)^2$ .

When $x=1$ , this is equal to

$2f''(2)+3f'(2)^2$

We know $f''(2)=0$ , since $f'$ has a horizontal tangent at $x=2$ , and from the graph of $f'$ we also know $f'(2)=5$ . So we find $h''(1)=75>0$ , which means $h$ is concave upward at the point (1, 2).

(e) $h$ is increasing when $h'>0$ . The sign of $h'$ is determined by $f'(2x)$ . We're interested in the interval $-1$ , and the behavior of $h$ on this interval depends on the behavior of $f'$ on $-2$ .