[tex]g(x)=e^{ax}+f(x)\implies g'(x)=ae^{ax}+f'(x)\implies g''(x)=a^2e^{ax}+f''(x)[/tex]
Given that [tex]f'(0)=5[/tex] and [tex]f''(0)=7[/tex], it follows that
[tex]g'(0)=a+5[/tex]
[tex]g''(0)=a^2+7[/tex]
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[tex]h(x)=\cos(kx)f(x)+\sin x\implies h'(x)=-k\sin(kx)f(x)+\cos(kx)f'(x)+\cos x[/tex]
When [tex]x=0[/tex], we have
[tex]h(0)=\cos0f(0)+\sin0=f(0)=3[/tex]
The slope of the line tangent to [tex]h(x)[/tex] at (0, 3) has slope [tex]h'(0)[/tex],
[tex]h'(0)=-k\sin0f(0)+\cos0f'(0)+\cos0=5+1=6[/tex]
Then the tangent line at this point has equation
[tex]y-3=6(x-0)\implies y=6x+3[/tex]
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Differentiating both sides of
[tex]4x^2+y^2=48+2xy[/tex]
with respect to [tex]x[/tex] yields
[tex]8x+2y\dfrac{\mathrm dy}{\mathrm dx}=2y+2x\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]\implies(2y-2x)\dfrac{\mathrm dy}{\mathrm dx}=2y-8x[/tex]
[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y-4x}{y-x}[/tex]
On this curve, when [tex]x=2[/tex] we have
[tex]4(2)^2+y^2=48+2(2)y\implies y^2-4y-32=(y-8)(y+4)=0\implies y=8[/tex]
(ignoring the negative solution because we don't care about it)
The tangent to this curve at the point [tex](x,y)[/tex] has slope [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex]. This tangent line is horizontal when its slope is 0. This happens for
[tex]\dfrac{y-4x}{y-x}=0\implies y-4x=0\implies y=4x[/tex]
and when [tex]x=2[/tex], there is a horizontal tangent line to the curve at the point (2, 8).
