calc problem?

Question

let f be a differentiable function with f(2) = 3 and f'(2) = -5, and let g be the function defined by g(x) = x. Which of the following is an equation of the line tangent to the graph of g at the point where x = 2.

would anyone care to lend me a hand?

let f be a differentiable function with f(2) = 3 and f '(2) = -5, and let g be the function defined by g(x) = xf(x). Which of the following is an equation of the line tangent to the graph of g at the point where x = 2:
a)y=3x
b)y-3=-5(x-2)
c)y-6=-5(x-2)
d)y-6=-7(x-2)
e)y-6=-10(x-2)

Answer 1

Yeah, something's missing there, I think.

If not, then I'd call it a trick question.

If g(x) = x, then the derivative, g'(x) = 1. So the equation of the "tangent line" at x = 2, would just be the line itself, y = x.

Answer 2

To find the tangent line of a function at a given point, first find its slope by calculating the value of the derivative there. With that information, you can figure out the intercept and have the line equation.

For your question, it seems like you haven't copied it exactly or something. The line tangent to g(x) = x is always y = x ...

Is there another f somewhere in the problem you left out?

Answer 3

D(x) = 20x + 10 = a hundred and ten, using fact the dam is a hundred thirty feet intense and 20 feet above the river line. remedy for x right here: 20x = a hundred, x = 5. W(x) = 10(x^2 - 8x + 22) = 310 feet, so x^2 - 8x + 22 = 31, set up as a quadratic equation subtracting 31 from the two aspects: x^2 - 8x - 9 = 0, factoring, (x+one million)(x-9) = 0, utilising the guideline: what 2 numbers such that their sum is -8 and product is -9? one million and -9. answer could desire to be useful, x = 9. place the dam the place x=5 for D(x) and x=9 for W(x). this would not make entire sense to me yet I a minimum of happy your equations.

Answer 4

Since f(2)=3 and f(2) = -5, f(x) is not a function. To be a function, there must be only one value of y for each value of x. Have you stated the problem correctly?

Glad to help you if you resubmit problem.

Answer 5

I'm assuming you're asking about the tangent line to f, since g is a line.

You can use the point-slope form:
y - y_0 = m ( x - x_0 )
The slope, m, is another way of writing f'(x_0). In your problem, x_0 is 2, y_0=f(x_0)=3 and m=f'(x_0)=-5.
The equation is y - 3 = (-5)( x - 2)

Answer 6

-5