Tangent line question to a power function

Answer 1

Mike

The slope of a line can be found by differentiating y

y'=(7)*(10x+7)*e^{5x^2 + 7x -1}

We know y'=-21*e^-3

(7)*(10x+7)*e^{5x^2 + 7x - 1} = -21*e^-3

Consider two equations:

(7)*(10x+7) = -21

e^{5x^2 + 7x - 1} = e^-3

7*(10x+7) = -21 means x = -1

e^{5x^2 + 7x - 1} = e^-3 means 5x^2+7x-1 = -3 or 5x²+7x+2=0 which means x=-1 or x=-2/5

Since x = -1 is a common solution to both equation, x = -1 is the unique solution.

The equation of a tangent line with slope m at point (a,b) is y-a = m* (x -a). See if you can find the equation of the tangent line form this hint.

Answer 2

Remember, slope of a tangent line ⇒ derivative. Therefore,

dy/dx = 7e^(5x² + 7x - 1)*d/dx(5x² + 7x - 1) (Remember your chain rule.) = 7(10x + 7)*e^(5x² + 7x - 1)

*Make note: The derivative of e^(x) = e^(x) REGARDLESS of what the exponent is, the derivative is just the same thing. The only thing you need to do is worry about your chain rule (when necessary). So to clarify:

d/dx (e^(9x+1)) = e^(9x+1)*d/dx(9x + 1) = 9e^(9x+1)

d/dx(e^(sinx)) = e^(sinx)*d/dx(sinx) = cos(x)*e^sin(x)

d/dx(e^(1/x)) = d/dx(e^{(x^(-1))}) = e^{(x^(-1))}*d/dx(x^(-1)) = -1x^(-2)*e^{(x^(-1))}

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Okay, so back to your problem, it said your slope of the tangent was -21e^(-3), so dy/dx = -21e^(-3)

7(10x + 7)*e^{(5x^2+7x-1)} = -21e^(-3)........(1)

Hence,

7(10x + 7) = -21 ⇒ 10x + 7 = -3 ⇒ x = -1

(You can check, if you plug in x = -1 into (1) the left side will equal the right side)

All you need now is the y-value at x = -1. So, go back to your original equation, and substitute x = -1

y = 7e^{(5(-1)^2 +7(-1) -1)} = 7e^-3

Hence, the equation of your tangent line is:

y - 7e^-3 = -21e^-3(x + 1)

Hope this is helpful.

Mr. K