Find tangent line equation

Question

Find the equation of the line tangent to y = f(x) at x = 1, where

f(x) = 7e^4x + e^{-x^2}

all calculations must be exact and round the final answers to two decimal places.

the equation of the tangent line is y = mx+b, where

m =

b =

Answer 1

So, you have a given function f(x), and you want to find the equation of the line tangent to the point (1, f(1)). If you substitute 1 into your function, you obtain 7e⁴+e^-1. So, your point is (1, 7e⁴+e^-1). For your equation y=mx+b, you now have x=1 and y=7e⁴+e^-1.

Next, you want to find m, the slope. Take the derivative of your function f(x)=7e^4x + e^{-x^2}.

You need to use the chain rule to do this. We are dealing with powers of e. To take the derivative, you must first take the derivative of whatever is within the exponent, and then multiply that by your original exponent.

For e^4x, if you focus only on the 4x and take the derivative, that is 4. Multiply this result by your original function, e^4x, to obtain 4e^4x.

For e^{-x^2}, if you focus only on the -x^2 and take the derivative, that is -2x. Multiply this result by your original function, e^{-x^2}, to obtain -2xe^{-x^2}.

Put these derivatives back into the original function as follows:

f ' (x) = 7(4e^4x) -2xe^{-x^2} = 28e^4x-2xe^{-x^2}

Now to find the value of the slope m at f(1), we plug 1 into our new equation. This gives us:

m=f ' (1) = 28e⁴-2e^-1

Since we now have values for x, y, and m, we can solve for b in the equation y=mx+b, as follows:

Start with y=mx+b, then substitute your values

7e⁴+e^-1=(28e⁴-2e^-1)(1)+b

7e⁴+e^-1=28e⁴-2e^-1+b

Subtract 28e⁴-2e^-1 from both sides to obtain:

b = -21e⁴+3e^-1

So, since you have your values, m=28e⁴-2e^-1 and b=b = -21e⁴+3e^-1, you may put each of these expressions into a calculator, which should give you m = 1528.01 and b = —1145.45 as your final answers. Your equation would be:

y=1528.01x-1145.45

Answer 2

The slope of the tangent line to f(x) at x=1, is just f’(1).

We have,

f’(x) = 28e^4x - 2xe^{-x^2}, from the chain rule.

f’(1) = 28e⁴ - (2/e) = slope of the tangent line at x = 1.

To find the equation of the tangent line, we want to use the point-slope formula. So we seek to find the y value that corresponds to x = 1.

evaluating f(x) at x = 1 yields f(1)= 7e⁴ - (1/e).

thus, y - ( 7e⁴ - (1/e)) =(28e⁴ - (2/e))(x-1).

I’ll leave it to you to to put it into slope-intercept form (y=mx + b).