The line tangent to the curve y=h(x) at x=4 is y = −3x+14. Fin...

Question

The line tangent to the curve y=h(x) at x=4 is y = −3x+14. Find an equation of the line tangent to the following curves at x=4.

y = (x²-3x)h(x)

Accepted Answer

Welcome back, everyone. In this problem, given that the tangent line to the curve Y equals F of X at X equals 3 is Y equals -2 X + 5, determine the equation of the tangent line to the curve Y equals 3X minus X2red multiplied by F X at X equals 3. A says the equation is Y equals 3X minus 9, B Y equals 3 X + 9, C Y equals 3 x + 3, and the D Y equals -3 X minus 3. Now, if we're going to determine the equation of the tangent line to the curve Y equals 3X minus X2 multiplied by F of X at X equals 3, 1st, we'll need to get the slope and the point. Recall that by the point slope form of an equation, we can rewrite the equation of a straight line as Y minus Y1 equals M multiplied by X minus X1. Now, if we're going to write our equation in this form, first, we'll need the slope M, and then we'll need the points X1 and Y1. So how can we find the slope of our tangent line? Well, we can differentiate or curve Y at the point where X equals 3. Now notice here that Y is a product. 3X minus X2red multiplied by F of X. Now, since it's a product, we can differentiate it using the po rule. So let's find the derivative of Y of X, then solve for it at X equals 3, and then use that along with the points X1 and Y1 to find the equation of the tangent line. Now, what do we know about differentiating with the product tool? If we differentiate by the product rule, OK, it basically tells us that if we have a function of XY equal UV where U and V are functions of X. Then the derivative of y with respect to X equals the derivative of u multiplied by V plus the derivative of V multiplied by U. In other words, we can differentiate both terms and then multiply by the opposite original function and then simplify. So in this problem, let's let you be equal to 3X minus X squad. And we. B equal to F of X. Then that means U is equal to 3 minus 2X. When we differentiate 3 X is 3, derivative of X squared is 2 X. And the derivative of V is going to be the derivative of F of X, OK? Now, remember, we know that F of X, Y equals F of X for the curve, and the tangent line to the curve is Y equals -2 X + 5. So when we differentiate F of X, we get it to be -2. Now that we have U and U and V, we can figure out what they will be at X equals 3. So now at X equals 3, then U, which is 3X minus X2, is gonna be 3 multiplied by 3 minus 32, which is 32 minus 320 and U is 3 minus 2 multiplied by 3. That's 3 minus 6, which is negative3. Next, we know that V is FF3 and F of 3, we know F of X is -2 X + 5. So FF 3 is -2 multiplied by 3 + 5, which equals -1. And already we know that the prime. I'll just fix that here. The derivative of V will be the derivative of F of 3, OK, which we already found to be a constant negative 2. Know that we have our values, OK, for U, U, V and V when X is equal to 3, then we can find a derivative. Of why When X is 3, OK. I should put that in red actually. So now that derivative, the derivative of Y and X is 3 is going to be equal to -3 U multiplied by V-1 plus. U 0 multiplied by V-2. That just gives us -3 multiplied by -1, which is 3. So now we have our slope M, which is equal to 3. OK. So, the derivative of Y at X equals 3 is our slope. Next, what about our points? Well, technically we, we already found it because we know, OK, that when X equals 3, Y equals 0, so we have a point on the line. Thus we can use the point slope form of our equation. So now, this means we can write our equation as Y minus Y1, which is 0, is going to be equal to or slope 3 multiplied by X minus X1, which is 3. Now if you think about it, Y minus 0 is 0 and 3 multiplied by X minus 3 is 3X minus 9. Therefore, Y is equal to 3X minus 9. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

The line tangent to the curve y=h(x) at x=4 is y = −3x+14. Find an equation of the line tangent to the following curves at x=4.y = (x²-3x)h(x)

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The line tangent to the curve y=h(x) at x=4 is y = −3x+14. Find an equation of the line tangent to the following curves at x=4.
y = (x²-3x)h(x)