thanks!Find the value of c for which the line $\displaystyle y=x+c$ is a tangent to the curve $\displaystyle y=x^2-5x+4$
If it is tangent to the curve, it will touch the curve only once.
Since they touch, they must be equal.
So $\displaystyle x + c = x^2 - 5x + 4$
$\displaystyle x^2 - 6x + 4 - c = 0$
Since it only touches once, the discriminant must be 0.
So $\displaystyle \Delta = (-6)^2 - 4(1)(4 - c) = 0$
$\displaystyle 36 - 16 + 4c = 0$
$\displaystyle 20 + 4c = 0$
$\displaystyle 4c = -20$
$\displaystyle c = -5$.
This was posted in the "Pre Calculus" section so Prove It used a method that did not require the derivative.
Since you do mention the derivative, no, the derivative is not 0 where the line y= x+ c is tangent to it. A line is tangent to a curve where its slope is the same as the derivative. y= x+ c has slope 1 so you are looking for a place where the derivative is 1, not 0.
Here is a method using the derivative.
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The derivative of the function $\displaystyle f(x) = x^2 - 5x + 4$ is $\displaystyle f'(x) = 2x - 5$ (power rule on sum of functions).
The line $\displaystyle y = x + c$ has a slope equal to $\displaystyle 1$. Thus, you are looking for the point on the curve of $\displaystyle f(x)$ where the slope is equal to $\displaystyle 1$. So, you must solve $\displaystyle 2x - 5 = 1$ for $\displaystyle x$. Hmm, $\displaystyle x = 3$.
Say $\displaystyle a = 3$ (to make it less confusing). You know that the tangent to the curve of $\displaystyle f$ in a point of absciss $\displaystyle a$ is equal to :
$\displaystyle y = (x - a)f'(a) + f(a)$
Substitute : $\displaystyle y = (x - 3) \times 1 - 2$
That is : $\displaystyle y = x - 5$. Therefore $\displaystyle c = -5$.