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# Let $g(x) = e^{\alpha} + f(x)$ and $h(x) = e^{kx},$ where $f(0) = 3, f'(0) = 5,$ and $f" (0) = -2.$(a) Find $g'(0)$ and $g"(0)$ in terms of $c$. b) In terms of $k,$ find an equation of the tangent line to the graph of $h$ at the point where $x = 0.$

## a. $g^{\prime}(0)=c+5$$g^{\prime \prime}(0)=c^{2}-2$b. $y=(3 k+5) x+3$

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were given function G of X and we want to find g prime of zero N g double prime A zero. So let's find G prime of X. So using the chain rule on each of the C X, we get each of the C X times see, and then the derivative of F of X would be f prime of X. Now let's evaluate that a zero so g prime a zero would be each of the sea time zero time C plus f prime of zero now e to the C times zero is each of the zero and each of the zero is one. So we have one time see plus F Prime of zero and F Prime of zero is five. So if c plus five so g prime of zero equals C plus five. Okay, Now let's find G double prime of zero, starting with G double prime of X. So we're going to find the derivative of G Prime of X. So we have see the constant, and then we have the derivative of E to the c X. We're looking at this right here. The derivative of each of the C X would be e to the C X time. See, and then we have the derivative of F prime of X, and that would be F double prime of X. Let's evaluate this at zero. So G double prime of zero would be see Time's each of the zero time see plus F double prime of zero Eat of zero again is once we have C times one time, see so that C squared and then f double prime of zero is negative. Two. We've C squared minus two and that's G double Prime A zero. Okay, now, part be it. The problem is working with h of X and we're going to find the equation of the tangent line at X equals zero. So we need to find aged prime of X because that's the derivative, and the slope of the tangent line will be the derivative at that point. So age prime of X, we're going to need to use the product rule. So we have the first each of the k x times, the derivative of the second F prime of X, plus the second f of x times, a derivative of the first E to the K X Times K. Now let's substitute zero in there because we want the slope at X equals zero. So of age Prime of zero equals e to the K time zero that would be each of the zero times F prime of zero plus f of zero times each of the K time. Zero that would be eating the zero times K. All right. F Prime of zero is five, so we can substitute that in and effort zero is three, so we can substitute that in eat of the zero is one. So we have one times five class, three times one times k. So we have five plus three K. That's the slope of the tangent line. Now we need the equation of the Tanja line. So let's make a little room here now. We can use why minus y one equals X Times X minus RM times X minus X one That would be the point slope form of the equation of a line we know the X coordinate ever point is zero. What's the Y. Coordinate of our point? What's H of zero each of the K time zero plus f of zero? That would be one plus an f of zero that I just erased was 31 plus three. That's four. So this is the 40.4 so we can use that as our X one. And why, once we have y minus four equals Air Slope, which I'm going to write as three K plus five times x minus x one x minus zero So this is why I minus four, equals the quantity three K plus five Times X, and then we can add four to both sides and we have y equals three K plus five times X plus four. Okay, Did you catch my error earlier? I put a plus sign in where I shouldn't have. So it's e to the zero times f of zero finding h of zero. So that's actually three, so we can easily fix that. Just go back and change our for two or three. No problem. All the same concept supply, and we're going to get the equation of the tangent line. Why? He calls 33 K plus five times X plus three. There we go

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