Question

The twice–differentiable function f is defined for all real numbers and satisfies the following conditions: f(0)=3 f′(0)=5 f″(0)=7 The function g is given by g(x)=eax+f(x) for all real numbers, where a is a constant. Find g ′(0) and g ″(0) in terms of a. Show the work that leads to your answers. The function h is given by h(x)=cos(kx)[f(x)]+sin(x) for all real numbers, where k is a constant. Find h ′(x) and write an equation for the line tangent to the graph of h at x=0. For the curve given by 4x2+y2=48+2xy show that dy dx = y−4x y−x . For the curve given by 4x2+y2=48+2xy, find the positive y-coordinate given that the x-coordinate is 2. For the curve given by 4x2+y2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

Answer 1

$g(x)=e^{ax}+f(x)\implies g'(x)=ae^{ax}+f'(x)\implies g''(x)=a^2e^{ax}+f''(x)$

Given that $f'(0)=5$ and $f''(0)=7$, it follows that

$g'(0)=a+5$

$g''(0)=a^2+7$

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$h(x)=\cos(kx)f(x)+\sin x\implies h'(x)=-k\sin(kx)f(x)+\cos(kx)f'(x)+\cos x$

When $x=0$, we have

$h(0)=\cos0f(0)+\sin0=f(0)=3$

The slope of the line tangent to $h(x)$ at (0, 3) has slope $h'(0)$,

$h'(0)=-k\sin0f(0)+\cos0f'(0)+\cos0=5+1=6$

Then the tangent line at this point has equation

$y-3=6(x-0)\implies y=6x+3$

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Differentiating both sides of

$4x^2+y^2=48+2xy$

with respect to $x$ yields

$8x+2y\dfrac{\mathrm dy}{\mathrm dx}=2y+2x\dfrac{\mathrm dy}{\mathrm dx}$

$\implies(2y-2x)\dfrac{\mathrm dy}{\mathrm dx}=2y-8x$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y-4x}{y-x}$

On this curve, when $x=2$ we have

$4(2)^2+y^2=48+2(2)y\implies y^2-4y-32=(y-8)(y+4)=0\implies y=8$

(ignoring the negative solution because we don't care about it)

The tangent to this curve at the point $(x,y)$ has slope $\dfrac{\mathrm dy}{\mathrm dx}$. This tangent line is horizontal when its slope is 0. This happens for

$\dfrac{y-4x}{y-x}=0\implies y-4x=0\implies y=4x$

and when $x=2$, there is a horizontal tangent line to the curve at the point (2, 8).