Mathematics: Need help on both of these Calculus AB mock exams.

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Need help on both of these Calculus AB mock exams.

Ответ:

00bebebeauty00

25.01.2022

AB1

(a) is continuous at x=3 if the two-sided limit exists. By definition of , 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 is continuous at x=3 .

(b) Judging by the graph of , we know 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 , we see that $f'\ge0$ , meaning 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 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 , 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-10 , 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 has a horizontal tangent at x=2 , and from the graph of we also know f'(2)=5 . So we find h''(1)=750 , which means is concave upward at the point (1, 2).

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